# 14.3 Relative strengths of acids and bases  (Page 13/18)

 Page 13 / 18

From the equilibrium concentrations given, calculate K a for each of the weak acids and K b for each of the weak bases.

(a) CH 3 CO 2 H: $\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ = 1.34 $×$ 10 −3 M ;
$\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]$ = 1.34 $×$ 10 −3 M ;

[CH 3 CO 2 H] = 9.866 $×$ 10 −2 M ;

(b) ClO : [OH ] = 4.0 $×$ 10 −4 M ;

[HClO] = 2.38 $×$ 10 −5 M ;

[ClO ] = 0.273 M ;

(c) HCO 2 H: [HCO 2 H] = 0.524 M ;
$\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ = 9.8 $×$ 10 −3 M ;
$\left[{\text{HCO}}_{2}{}^{\text{−}}\right]$ = 9.8 $×$ 10 −3 M ;

(d) ${\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}:$ $\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right]$ = 0.233 M ;

[C 6 H 5 NH 2 ] = 2.3 $×$ 10 −3 M ;
$\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ = 2.3 $×$ 10 −3 M

From the equilibrium concentrations given, calculate K a for each of the weak acids and K b for each of the weak bases.

(a) NH 3 : [OH ] = 3.1 $×$ 10 −3 M ;
$\left[{\text{NH}}_{4}{}^{\text{+}}\right]$ = 3.1 $×$ 10 −3 M ;

[NH 3 ] = 0.533 M ;

(b) HNO 2 : $\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ = 0.011 M ;
$\left[{\text{NO}}_{2}{}^{\text{−}}\right]$ = 0.0438 M ;

[HNO 2 ] = 1.07 M ;

(c) (CH 3 ) 3 N: [(CH 3 ) 3 N] = 0.25 M ;
[(CH 3 ) 3 NH + ] = 4.3 $×$ 10 −3 M ;

[OH ] = 4.3 $×$ 10 −3 M ;

(d) ${\text{NH}}_{4}{}^{\text{+}}:$ $\left[{\text{NH}}_{4}{}^{\text{+}}\right]$ = 0.100 M ;

[NH 3 ] = 7.5 $×$ 10 −6 M ;
[H 3 O + ] = 7.5 $×$ 10 −6 M

(a) ${K}_{\text{b}}=1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5};$
(b) ${K}_{\text{a}}=4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4};$
(c) ${K}_{\text{b}}=7.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5};$
(d) ${K}_{\text{a}}=5.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}$

Determine K b for the nitrite ion, ${\text{NO}}_{2}{}^{\text{−}}.$ In a 0.10- M solution this base is 0.0015% ionized.

Determine K a for hydrogen sulfate ion, ${\text{HSO}}_{4}{}^{\text{−}}.$ In a 0.10- M solution the acid is 29% ionized.

${K}_{\text{a}}=1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}$

Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:

(a) F

(b) ${\text{NH}}_{4}{}^{\text{+}}$

(c) ${\text{AsO}}_{4}{}^{3-}$

(d) ${\left({\text{CH}}_{3}\right)}_{2}{\text{NH}}_{2}{}^{\text{+}}$

(e) ${\text{NO}}_{2}{}^{\text{−}}$

(f) ${\text{HC}}_{2}{\text{O}}_{4}{}^{\text{−}}$ (as a base)

Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:

(a) HTe (as a base)

(b) ${\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{\text{+}}$

(c) ${\text{HAsO}}_{4}{}^{3-}$ (as a base)

(d) ${\text{HO}}_{2}{}^{\text{−}}$ (as a base)

(e) ${\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}$

(f) ${\text{HSO}}_{3}{}^{\text{−}}$ (as a base)

(a) ${K}_{\text{b}}=4.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-12};$
(b) ${K}_{\text{a}}=1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8};$
(c) ${K}_{\text{b}}=5.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7};$
(d) ${K}_{\text{b}}=4.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3};$
(e) ${K}_{\text{b}}=2.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3};$
(f) ${K}_{\text{b}}=6.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-13}$

For which of the following solutions must we consider the ionization of water when calculating the pH or pOH?

(a) 3 $×$ 10 −8 M HNO 3

(b) 0.10 g HCl in 1.0 L of solution

(c) 0.00080 g NaOH in 0.50 L of solution

(d) 1 $×$ 10 −7 M Ca(OH) 2

(e) 0.0245 M KNO 3

Even though both NH 3 and C 6 H 5 NH 2 are weak bases, NH 3 is a much stronger acid than C 6 H 5 NH 2 . Which of the following is correct at equilibrium for a solution that is initially 0.10 M in NH 3 and 0.10 M in C 6 H 5 NH 2 ?

(a) $\left[{\text{OH}}^{\text{−}}\right]=\left[{\text{NH}}_{4}{}^{\text{+}}\right]$

(b) $\left[{\text{NH}}_{4}{}^{\text{+}}\right]=\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right]$

(c) $\left[{\text{OH}}^{\text{−}}\right]=\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right]$

(d) [NH 3 ] = [C 6 H 5 NH 2 ]

(e) both a and b are correct

(a) is the correct statement.

Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.25 M in HCO 2 H and 0.10 M in HClO.

Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.134 M in HNO 2 and 0.120 M in HBrO.

[H 3 O + ] = 7.5 $×$ 10 −3 M
[HNO 2 ] = 0.127
[OH ] = 1.3 $×$ 10 −12 M
[BrO ] = 4.5 $×$ 10 −8 M
[HBrO] = 0.120 M

Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.25 M in CH 3 NH 2 and 0.10 M in C 5 H 5 N ( K b = 1.7 $×$ 10 −9 ).

Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.115 M in NH 3 and 0.100 M in C 6 H 5 NH 2 .

[OH ] = $\left[{\text{NO}}_{4}{}^{\text{+}}\right]$ = 0.0014 M
[NH 3 ] = 0.144 M
[H 3 O + ] = 6.9 $×$ 10 −12 M
$\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right]$ = 3.9 $×$ 10 −8 M
[C 6 H 5 NH 2 ] = 0.100 M

Using the K a value of 1.4 $×$ 10 −5 , place $\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}$ in the correct location in [link] .

Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants can be found in Appendix H and Appendix I .

(a) 0.0092 M HClO, a weak acid

(b) 0.0784 M C 6 H 5 NH 2 , a weak base

(c) 0.0810 M HCN, a weak acid

(d) 0.11 M (CH 3 ) 3 N, a weak base

(e) 0.120 M $\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{2+}$ a weak acid, K a = 1.6 $×$ 10 −7

(a) $\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{ClO}}^{\text{−}}\right]}{\left[\text{HClO}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{\left(0.0092-x\right)}\phantom{\rule{0.2em}{0ex}}\approx \phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{0.0092}\phantom{\rule{0.2em}{0ex}}=2.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}$
Solving for x gives 1.63 $×$ 10 −5 M . This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H 3 O + ] = [ClO]= 5.8 $×$ 10 −5 M
[HClO] = 0.00092 M
[OH ] = 6.1 $×$ 10 −10 M ;
(b) $\frac{\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right]\left[{\text{OH}}^{\text{−}}\right]}{\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{\left(0.0784-x\right)}\phantom{\rule{0.2em}{0ex}}\approx \phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{0.0784}\phantom{\rule{0.2em}{0ex}}=4.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}$
Solving for x gives 5.81 $×$ 10 −6 M . This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
$\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]$ = [OH ] = 5.8 $×$ 10 −6 M
[C 6 H 5 NH 2 ] = 0.00784
[H 3 O + ] = 1.7 $×$ 10 −9 M ;
(c) $\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{CN}}^{\text{−}}\right]}{\left[\text{HCN}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{\left(0.0810-x\right)}\phantom{\rule{0.2em}{0ex}}\approx \phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{0.0810}\phantom{\rule{0.2em}{0ex}}=4.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}$
Solving for x gives 6.30 $×$ 10 −6 M . This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H 3 O + ] = [CN ] = 6.3 $×$ 10 −6 M
[HCN] = 0.0810 M
[OH ] = 1.6 $×$ 10 −9 M ;
(d) $\frac{\left[{\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{\text{+}}\right]\left[{\text{OH}}^{\text{−}}\right]}{\left[{\left({\text{CH}}_{3}\right)}_{3}\text{N}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{\left(0.11-x\right)}\phantom{\rule{0.2em}{0ex}}\approx \phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{0.11}\phantom{\rule{0.2em}{0ex}}=6.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$
Solving for x gives 2.63 $×$ 10 −3 M . This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[(CH 3 ) 3 NH + ] = [OH ] = 2.6 $×$ 10 −3 M
[(CH 3 ) 3 N] = 0.11 M
[H 3 O + ] = 3.8 $×$ 10 −12 M ;
(e) $\frac{\left[\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left({\text{OH}\right)}^{\text{+}}\right]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{\left[\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{2+}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{\left(0.120-x\right)}\phantom{\rule{0.2em}{0ex}}\approx \phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{0.120}\phantom{\rule{0.2em}{0ex}}=1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}$
Solving for x gives 1.39 $×$ 10 −4 M . This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[Fe(H 2 O) 5 (OH) + ] = [H 3 O + ] = 1.4 $×$ 10 −4 M
$\left[\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{2+}\right]$ = 0.120 M
[OH ] = 7.2 $×$ 10 −11 M

Propionic acid, C 2 H 5 CO 2 H ( K a = 1.34 $×$ 10 −5 ), is used in the manufacture of calcium propionate, a food preservative. What is the hydronium ion concentration in a 0.698- M solution of C 2 H 5 CO 2 H?

White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g/cm 3 , what is the pH?

pH = 2.41

The ionization constant of lactic acid, CH 3 CH(OH)CO 2 H, an acid found in the blood after strenuous exercise, is 1.36 $×$ 10 −4 . If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution?

Nicotine, C 10 H 14 N 2 , is a base that will accept two protons ( K 1 = 7 $×$ 10 −7 , K 2 = 1.4 $×$ 10 −11 ). What is the concentration of each species present in a 0.050- M solution of nicotine?

[C 10 H 14 N 2 ] = 0.049 M
[C 10 H 14 N 2 H + ] = 1.9 $×$ 10 −4 M
$\left[{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}_{2}{}^{2+}\right]$ = 1.4 $×$ 10 −11 M
[OH ] = 1.9 $×$ 10 −4 M
[H 3 O + ] = 5.3 $×$ 10 −11 M

The pH of a 0.20- M solution of HF is 1.92. Determine K a for HF from these data.

The pH of a 0.15- M solution of ${\text{HSO}}_{4}{}^{\text{−}}$ is 1.43. Determine K a for ${\text{HSO}}_{4}{}^{\text{−}}$ from these data.

${K}_{\text{a}}=1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}$

The pH of a 0.10- M solution of caffeine is 11.16. Determine K b for caffeine from these data:
${\text{C}}_{8}{\text{H}}_{10}{\text{N}}_{4}{\text{O}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{10}{\text{N}}_{4}{\text{O}}_{2}{\text{H}}^{\text{+}}\left(aq\right)+{\text{OH}}^{\text{−}}\left(aq\right)$

The pH of a solution of household ammonia, a 0.950 M solution of NH 3, is 11.612. Determine K b for NH 3 from these data.

${K}_{\text{b}}=1.77\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$

#### Questions & Answers

if I change a wood into a book,is that called a chemical property?
does any one have an answer to the question
Bertram
the answer is yes since oxygen react with wood
Prax
no because the chemical composition of the wood remains the same
kabu
no
Vicky
Hi.... so it's physical property or?
Amandla
What is the difference between evaporation and boiling
how is it possible to make the volume of a gas zero?
what are the raw materials used during contact process
special Proteins who serve a Function as a Katalysator for chemical and biological reations
Sch
what is Enzymes
special Proteins who serve a Function as a Katalysator for chemical and biological reations
Sch
Does Ethers has hydrogen bonding with water?
ethers lack the hydroxyl groups of alcohols. when ether is placed in water both hydrogen atoms are replaced by alkyl or aryl groups. because ether lacks the O-H bond ether molecules cannot engage in hydrogen bonding with each other.
Jallal
however ethers do have nonbonding electron pairs on their oxygen atoms which allows them to form hydrogen bonds with other molecules such as alcohol raise or a amines that have O-H or N-H bonds.
Jallal
such as alcohols or amines* I always have. typo problem. I need to proofread before I send
Jallal
please what is the a difference between a compound and molecular formula
A compound is just a combination of different element showing their molecular formulation like h2o... A molecular formula is a formular showing the number if moles of each element in a compound
Its
a compound is made up of different atoms help together by ionic bonds in a fixed ratio. example formula is: NaCl or salt. a moleculer compound is consists of two or more atoms held together by covalent bonds. example formula is: H2, O2. the atoms can be the same or different such as CO2 or PCl5
Jallal
Yes, exactly what Its said is correct
Jallal
explain why the number of electron in the shells of elements making up the period of the periodic table following the pattern 2,8,18,32,50
these shells of elections emanating from the core on outward can hold only a certain amount of elections. the Pauli Exclusion rule states that no two elections can occupy the same space. an atom will always to fill all it's shells to maximum with the outer most shell called the valence shell.
Jallal
the first shell closet to the nucleus can hold only two elections. the second shell can hold 8 elections. the third shell can hold 18 elections. the fourth shell can hold 32 elections. the fifth shell can hold 50 elections. point people the atom will try to fully complete it's outer shells
Jallal
this cannot always be done though. the final outermost orbital she'll is called the valence shell and that is what determines it's reactivity. Hence the reason they're placed on the periodic table in that order
Jallal
Noble elements on the furthest right hand side of the periodic table of elements are also called inert because they don't react to other elements because they have all that shells already filled
Jallal
the valence shell also determines the chemicals properties of the atom.
Jallal
The general formula is that the (n)th shell can in principle hold up to 2(n^2) elections.
Jallal
what kind of force firmly hold the electrons on its orbit?
Techiman
The ISBN is going through
the ISBN is either a 9 digit number or 13 digit number. all textbooks have both those numbers. it's not a big deal anyway
Jallal
the ISBN didn't work. try looking for the 13 digit number instead
Jallal
thanks jallal
Faith
but pls am having problem with solubility curve
Faith
could you please detail what you're having trouble on exactly regarding solubility?
Jallal
how to find oxidation number of an element
Faith
for instance if you are given a graph to plot and answer the question below is kind of confusing
Faith
if there supposed to be a graph?
Jallal
write out the question to the problem
Jallal
yes
Faith
it's not showing.
Jallal
just write out the exact question
Jallal
u know wat just explain oxidation number seems like solubility curve is meant to be self explanatory
Faith
ok first the basic rules for oxidation
Jallal
1.) atoms in their elemental state have an oxidation number of zero. 2.) atoms in monatomic (single atom) ions have an oxidation number equal to their charge
Jallal
3.) in compounds: fluorine is assigned a -1 oxidation number, oxygen is usually assigned a -2 oxidation number (except in peroxide compounds where it is -1, and in binary compounds with fluorine where it is positive); and hydrogen is usually assigned a +1 oxidation number except when it exists
Jallal
okay
Faith
as the hydride ion, H^-,
Jallal
4.) in compounds, all other atoms are assigned an oxidation number so that the sum of the oxidation numbers on all the atoms in the species equals a charge on the species
Jallal
now let's try to determine oxidation state of H2 and H2O
Jallal
using rule 1 H2 is already in its elemental state so it's oxidation number is 0
Jallal
for H2O we we must get the sum of all the oxidation numbers using formula (oxidation # of O x # of O atoms) + (oxidation # of H x # of H atoms)
Jallal
already known oxidation number of oxygen and hydrogen the formula will be (-2x1)+(+1x2)=0
Jallal
0 is exactly what we would expect for a neutral molecule and water is neutral.
Jallal
would you like another example?
Jallal
in dis kind of question that says that what is the oxidation number of z in k3 zcl6 how am I to do it
Faith
k has an oxidation number of +1 and Cl is -1. this is because K has one unpaired electron and its outermost shell and is electropositive being a metal while Cl has one unpaired electron in its outermost shell and is electronegative being nonmetal.
Jallal
so to solve it: k3zcl6.3 times 1 + z + -1 times 6=0. 3-6 +z=0. -3+z=0 so z=3
Jallal
every element in the periodic table has an oxidation number unless it's ionized
Jallal
that is how you find those values and work from there
Jallal
thanks alot
Faith
excuse me I meant to write z=+3
Jallal
the oxidation number has to be positive or negative so you must indicate that
Jallal
I hope I was clear enough. also I want to help you with the solubility problem. if you can tell me the question of the problem I can find a solution and help explain it
Jallal
remember to always use the equation I wrote out before to get the sum of oxidation numbers when dealing with compounds
Jallal
What is collision theory
why would you not make potassium chloride from potassium and hydro caloric acids
If HCl and K were to react to form KCl, then they would have to liberate hydrogen gas, but because HCl is an acid it liberates hydrogen gas only if it reacts with very electropositive metals
Rahim
what is the hardest substance
with what scale one would be 💎
coland
diamond a 10
coland
1-10
coland
there is actually other elements that can cut diamonds and also withstand greater hear and temperature than diamonds.
Jallal
What ate the elements?, Jallal
Rahim
Sorry "ate" should be "are".Typing error.
Rahim
Pls Jallal ,can you name those elements because it is only diamond that I know to be the hardest substance for now.
hamidat
I thought I replied to this already. My apologies for the delay. Diamonds while very hard are not the hardest material on Earth. There is boron nitride but out of the 4 forms it can take it is the Wurtzite form (w-BN) that is harder than diamond.
Jallal
Next is Lonsdaleite which is also called hexagonal diamond because of its crystal structure but it is an allotrope of carbon with a hexagonal lattice. the lattice structure is packed even more closely than that of diamonds making Lonsdaleite 58% harder than diamond
Jallal
Then there's fullerene which is also an allotrope of carbon whose molecule consists of carbon atoms connected by single and double bonds so as to form a closed or partially closed mesh was fused rings of 5 to 7 atoms
Jallal
the molecule may be a hollow sphere an ellipsoid a tube are many other shapes and sizes. graphene which is a flat mesh of regular hexagonal rings can be seen as an extreme member of the family. because of their typical soccer balls like shape there are more commonly referred to as buckyballs.
Jallal
fullerenes which were discovered by accident greatly expanded known allotropes of carbon. this discovery gave rise to carbon nanotubes which are even harder than diamonds. each nanotubes is between 2 and 4 nanometers yeah it is incredibly strong and tough. weighing only 10% of the weight of steel
Jallal
but has hundreds of times the strength of steel
Jallal
then there is graphing which is a hexagonal carbon lattice that's only a single atom thick which is arguably the most revolutionary material to be developed and utilize in 21st century. the basic structural element of graphene are carbon nanotubes. in proportion to its thickness it is the strongest
Jallal
material known
Jallal
then there is beta carbon nitride which is a super hard material predicted to to have a hardness equal or above diamond.
Jallal
Thanks,but I have to go and research more to know better.
hamidat
lastly there is linear acetylenic carbon also called carbyne which is also an allotrope of carbon that has a chemical structure as a repeating chain with alternating single and triple bonds. it would thus be the ultimate member of the polyyne family.
Jallal
T
hamidat
the this polymeric carbyne is a considerable interest in Android technology as its Young's modulus is 32.7 TPa which is 40 times that of a diamond.
Jallal
I thought amongst the allotropes of carbon,diamond is octahedral and graphite is hexagonal?
hamidat
carbyne chains have been claimed to be the strongest material known for density. calculations indicate that carbyne's specific tensile strength beats graphene, carbon nanotubes, and diamond.
Jallal
There you have it Rahim
Jallal
So far so good, you have mentioned about 6 to 7 other compounds that are harder than diamond, are these real?
hamidat
yes they are. I would never make up lies just to sound smart. please check them yourself as it's never good to just take someone's word without doing your own research
Jallal
there are other materials too but they are disputable and upon further testing on them they proved to fall short
Jallal
No that's not what I mean.
hamidat
Thank you for your time.
hamidat
the allotropes of carbon, diamond is tetrahedral lattice and graphite is bonded l together in sheets of a hexagonal lattice.
Jallal
I'm happy to help anytime I can. also what did you mean that's not what you meant? I'd like to fully answer your questions to your content
Jallal
Now I no better, thanks.
hamidat
you're welcome and anytime
Jallal
Sorry it was not meant for you,I was answering another question .
hamidat
okay Sir
hamidat
I can't reply through the other side,it filled up
hamidat
you can just create a new thread
Jallal
Diamond
Its
Diamond, while still one of the hardest substances, it has lost its top spot thanks to new discoveries
Jallal
I wish someone had more questions for me to answer. I like helping others and I love knowledge. Please everyone ask away no matter how trivial. Any question is always a good question. it's much better to be inquisitive than silent.
And I like asking people questions, so wait and collect yours.
hamidat
How can a mixture of petroleum be separated from kerosene?, which process is called the "Haber process", How is methane prepared laboratoryly.Thank u
hamidat
An organic compound decolorized acidified KMnO4 solution but failed to react with ammonical silver nitrate solution. the organic compound is likely to be?
Faith
hamidat. mixtures of two miscible liquids having a difference in their boiling points more than 25°C can be separated by the method of distillation.
Jallal
a mixture of kerosene in petroleum is taken in the distillation flask with a thermometer fit it in. you also need a beaker a water condenser in a Bunsen burner. once you have the apparatus set up then start to heat the mixture slowly. the thermometer should be watched simultaneously.
Jallal
kerosene will vaporize and condense in the water condenser. the condensed kerosene is collected from the condenser outlet whereas the petroleum is left behind and distillation flask.
Jallal
missing is prepared and laboratory by the protonation of methyl lithium and methyl magnesium iodide
Jallal
I'm not sure what you meant when you said which process is called the haber process. but I can say that the hub Haber-Bosch process is an artificial nitrogen fixation process and is the main industrial procedure for the production of ammonia today.
Jallal
the process converts atmospheric nitrogen (N2) to ammonia (NH3) buy a reaction with hydrogen (H2) using a metal catalyst under high temperatures and pressures.
Jallal
methane* is produced buy protonation of methyl lithium and muscle magnesium iodide. please excuse the typos
Jallal
ok I wanted to confirm because am having contradiction between fractional distillation and distillation itself.
hamidat
what's contradiction are you encountering
Jallal
ok well I'll always be happy to assist
Jallal
The reason why I ask for the Haber process is because some one told me is a process where hydrogen and nitrogen react to give ammonia and I was not sure.
hamidat
For methane, it is the laboratory preparation that am asking for.
hamidat
I see I gave you the laboratory synthesis but I think what you're looking for really is the industrial route synthesis. am I correct?
Jallal
You see,for distillation it has to do with two immiscible liquids so I normally misuse it for fractional distillation.
hamidat
The Haber process equation is : N2 + 3H2 ---> 2NH3 ∆H°=-91.8 kJ/mol
Jallal
Oh the one you gave above is the laboratory preparation/synthesis?
hamidat
yes, the one I gave was the laboratory synthesis not the industrial process I think that's what you we're looking for. I'll explain it right now
Jallal
Yes I have seen something like that: N2+3H2 _2NH3
hamidat
Sorry for the misunderstanding
hamidat
I only wish all the symbols could be displayed in the equation to avoid any confusion
Jallal
no worries. I like kitchen concepts
Jallal
Like seriously even I used to have that problem of symbols
hamidat
Methane (CH4) is one carbon in four atoms of hydrogen. it is a group 14 hydroid in the simplest alkane and is the main constituent of natural gas. through its relative abundance on Earth it is an attractive fuel however trying to capture and store it poses it's challenges due to its gaseous state
Jallal
under normal conditions for temperature and pressure.
Jallal
nothing is a tetrahedral molecule with four equivalent C-H bonds. it's electronic structure is described by four bonding molecular orbitals resulting from the overlap of the valence orbitals on C and H.
Jallal
it is also a component of petroleum gas right?
hamidat
at room temperature and standard pressure methane is a colorless odorless gas. The familiar smell of natural gas as used in homes is a chief with the addition of an odorant usually blends containing tert-butythoil as a safety measure.
Jallal
no it is a component of natural gas not petroleum gas
Jallal
OK
hamidat
nothing has a boiling point of -164 °C (-257.8 °F) at a pressure of 1 atmosphere. as a gas it is flammable over a range of concentrations (5.4-17%) in air at standard pressure.
Jallal
solid methane exist currently in 9 known modifications. cooling Messina normal pressure results in the formation of methane I. this substance crystallizes in the cubic system (space group Fm3m). the positions of the hydrogen atoms are not fixed in methane I. i.e. methane molecules may rotate freely.
Jallal
therefore it is a plastic Crystal
Jallal
the primary chemical reactions in methane or combustion, steam reforming to syngas and halogenation. in general medicine reactions are difficult to control.
Jallal
OK am with you
hamidat
Halogenation:addition of halogens right?
hamidat
yes halogenation is a chemical reaction that involves the addition of one or more halogens to a compound or material
Jallal
okay
hamidat
I hope you're right because I don't feel blessed but it's my nature to help in any way I can
Jallal
there are different types of halogenation of organic compounds by reaction type so you must be mindful of that. there are free radical halogenation, ketone halogenation, electrophilic halogenation, and halogen addition reaction. the structure of the substrate is one factor that determine the pathway
Jallal
okay now that we have all that information out of the way we are going to now move on to the industrial synthesis
Jallal
You don't talk like that because the Lord has so many ways of blessing people, you don't know what you have now until you lose it.So ways feel blessed okay
hamidat
You don't talk like that because the Lord has so many ways of blessing people, you don't know what you have now until you lose it.So always feel blessed okay
hamidat
okay
hamidat
all I have is my knowledge and my brain. I think about it everyday about what I have and while not much at all things could be magnitudes worse so you're right I should be happy with I currently have. maybe fate will smile down on me one day. thanks for the pep talk
Jallal
So are you still in secondary school preparing to write WAEC?
hamidat
no I'm finished with my university studies where I majored in the Sciences. I just love to keep expanding my knowledge
Jallal
No Wonder you sound different from normal school student
hamidat
now let's begin the industrial process. there is little incentive to produce methane industrially. methane is produced by hydrogenating carbon dioxide through the Sabatier process
Jallal
Alright
hamidat
the Sabatier process dissolve the reaction of hydrogen and carbon dioxide at elevated temperatures (optimally 300-400 °C) and pressures in the presence of a nickel catalyst to produce methane and water.
Jallal
okay
hamidat
Hello are you still there
hamidat
optimally ruthenium on alumina (aluminum oxide) make some more efficient catalyst. this process is described by the following exothermic reaction: CO2 + 4H2 ----> CH4 + 2H2O ∆H= -165.0 kJ/mol. above the arrow symbol there should be 400 °C and below it should say pressure
Jallal
sorry I had to talk to my family over something. don't worry I'll see this to the end and make sure you understand my explanation
Jallal
No problem sir,thanks for your time
hamidat
whether the CO2 methanation occurs by first associatively adsorbing an adatom hydrogen and forming oxygen intermediates before hydrogenation or dissociating and forming a carbonyl before being hydrogenated we are then given the next equation: CO + 3H2 ---> CH4 + H2O. ∆H= -206 kJ/mol
Jallal
CO methanation is believed to occur through a dissociative mechanism where the carbon oxygen bond is broken before hydrogenation with an associative mechanism only being observed at a high H2 concentrations
Jallal
messing is also a side product of the hydrogenation of carbon monoxide in the Fischer-Tropish process, which is practiced on a large scale to produce longer-chain molecules than methane
Jallal
the Fisher-Tropsch process involves a series of chemical reactions that produce a variety of hydrocarbons, ideally haven't formula (CnH2n+2). the more useful reactions produce alkanes as follows: (2n + 1) H2 + n CO ---> CnH2n+2 + n H2O
Jallal
okay but what do you mean by "The Fischer-Tropish" process
hamidat
where n is typically 10-20. the formation of methane (n=1) is unwanted. most of the alkanes produced tend to be straight-chain, suitable as diesel fuel. in addition to alkane formation, computing reactions give small amounts of alkenes, as well as alcohols and other oxygenated hydrocarbons.
Jallal
what I mean is that there are two ways to produce methane on an industrial scale. one is by hydrogenating carbon dioxide through the Sabatier process and the other as a side product of the Fischer-Tropsh process.
Jallal
as a side product of the hydrogenation of carbon monoxide in the Fisher-Tropsch process
Jallal
now I understand
hamidat
example of large-scale coal-to- methane gasification is the Great Plains Synfuels plant, in North Dakota as a way to develop abundant local resources of low-grade lignite, a resource that is otherwise difficult to transport for its weight, ash content, low calorific value and propensity
Jallal
just spontaneous combustion during storage and transport.
Jallal
power to methane is a technology that uses electrical power to produce hydrogen from water by electrolysis and uses the Sabatier reaction to combine hydrogen with carbon dioxide to produce methane.
Jallal
you see all these things are not in this secondary school textbooks that am using
hamidat
it depends on the books and authors of the books because I've seen academic books tremendously different from each other with completely different concepts
Jallal
also the edition and year of publication. my experience is from University. are yours from high school?
Jallal
it's through
hamidat
through?
Jallal
I'm almost done with my explanation. only about 5 sentences left to wrap it all up.
Jallal
Sorry what I meant is"true"
hamidat
okay sir
hamidat
also secondary school books compared to university school books offer much more information. they limit the amount of knowledge your exposed to deliberately so as you progress you learn more newer previously unknown concepts
Jallal
as of 2016 this is mostly under development not in large-scale use. theoretically, the process could be used as a buffer for excess and off-peak power generated by highly fluctuating wind generators and solar arrays.
Jallal
To be sincere, all what you have been teaching me is not there
hamidat
however, as currently very large amounts of natural gas are used in power plants to produce electric energy, the losses and efficiency are not acceptable.
Jallal
this ends the lesson
Jallal
you mean in your text book?
Jallal
Yes
hamidat
Faith. the answer is "an alkenes"
Jallal
could you please tell me the name of the book, author, and edition? if you can find the ISBN that would help me find it faster
Jallal
I answered. both your questions and Faith's. I went back to everything I wrote I apologise for all the typos
Jallal
Thanks so much Sir ,I have learnt a lot today. I really appreciate.
hamidat
you're always welcome. I enjoying teaching. if you ever have any other questions or need help please reach out anytime. although I'm not sure how one does that in this app
Jallal
good luck on your WAEC
Jallal
Essential chemistry, Odesina I .A.
hamidat
ISBN:978-8089-44-5.For edition he wrote first, second, third and fourth edition here .
hamidat
thanks Sir.
hamidat
my pleasure
Jallal
sir hope the ISBN appeared
hamidat
the ISBN didn't appear and it's proving difficult to find that book and author.
Jallal
just your the ISBN number
Jallal
I tried sending it but it won't go through
hamidat
it doesn't really matter regarder answer already through this whole topic thread
Jallal
Gibbs free energy continued
when ∆G>0, the process is endergonic and not spontaneous in the forward direction. instead it will proceed spontaneously in the reverse direction to make more starting materials
Jallal
when ∆G=0, the system is in equilibrium and the concentrations of the products and reactants will remain constant
Jallal
when the system is in equilibrium that means the forward reaction and the reverse reaction are occurring at the same rate
Jallal
Although ∆G is temperature dependent, it's generally okay to assume that the ∆H and ∆S values are independent of temperature as long as the reaction does not involve a phase change. that means that if we know ∆H and ∆S we can use those values to calculate ∆G at any temperature.
Jallal
calculating ∆H and ∆S can be done using tables of standard values among other methods
Jallal
when the process occurs under standard conditions (all gases at 1 bar pressure, all concentrations are 1 M, and T=25°C), we can also calculate ∆G using the standard free energy of formation, ∆fG°
Jallal
be sure to pay close attention two units when ∆G from ∆H and ∆S because ∆H is given in kJ/mol-reaction while ∆S is given as J/mol-reaction • K, which is a difference factor of 1000
Jallal
when is ∆G negative? the equation of ∆Gsystem depends on 3 values. using ∆Gsystem=∆Hsystem-T∆Ssystem, the temperature in this equation is always positive or zero because it has units of K there for the second term in our equation T∆Ssystem will always have the same sign as ∆Ssystem
Jallal
now we can make the following conclusions about when processes will have a negative ∆Gsystem
Jallal
when the process is exothermic (∆Hsystem < 0), and the entropy of the system increases (∆Ssystem > 0), the sign of ∆Gsystem is negative at all temperatures. thus the process is always spontaneous
Jallal
when the process is endothermic, ∆Hsystem > 0, and the entropy of the system decreases, ∆Ssystem < 0, the sign of ∆G is positive at all temperatures. thus the process is never spontaneous
Jallal
for other combinations of ∆Hsystem and ∆Ssystem, the spontaneity of a process depends on the temperature
Jallal
exothermic reactions (∆Hsystem < 0) that decrease the entropy of the system (∆Ssystem < 0) are spontaneous at low temperatures
Jallal
endothermic reactions (∆Hsystem > 0) The increased entropy of the system (∆Ssystem > 0) or spontaneous at high temperatures
Jallal
thermodynamics is also connected to concepts in other areas of chemistry for example, in chemical equilibrium we can relate ∆G with the equilibrium constant, K
Jallal
in electrochemistry, ∆G is related to the cell voltage, Ecell
Jallal
lastly depending on the signs of ∆H and ∆S, the spontaneity of a process can change at different temperatures
Jallal
if any part of this explanation of the concept of Gibbs free energy is not clear please let me know so I may clarify
Jallal
That is good my brother,continue.
hamidat
Jallal, I like the ways you answer questions, you always break it down to the simplest, I which I can get somebody like you that will be teaching me.
hamidat
I could have broken it down further by giving detailed sample problems but stopped short assuming everyone understood as I wasn't receiving feedback. if you'd like I'd be happy to teach you whenever you like or need. we could setup email correspondence or video chats
Jallal
OK whenever am ready I will let you know
hamidat
know what?
Jallal
I can skip to the very crucial process instead of giving a complete breakdown of how the process works
Jallal
You see am preparing for WAEC and I am a little nervous
hamidat
just keep studying and reviewing when you're not 100% on and go in feeling completely confident
Jallal
sorry for the typos
Jallal
let's continue
Jallal
Thank you so much,God will continue to bless you
hamidat
Why's Ph not good for consumption?