# 8.5 Hydrolysis of salt solutions  (Page 3/5)

 Page 3 / 5

## The ph of a solution of a salt of a weak base and a strong acid

Aniline is an amine that is used to manufacture dyes. It is isolated as aniline hydrochloride, $\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right]\text{Cl},$ a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of aniline hydrochloride?

${\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\left(aq\right)$

## Solution

The new step in this example is to determine K a for the ${\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}$ ion. The ${\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}$ ion is the conjugate acid of a weak base. The value of K a for this acid is not listed in Appendix H , but we can determine it from the value of K b for aniline, C 6 H 5 NH 2 , which is given as 4.3 $×$ 10 −10 ( [link] and Appendix I ):

${K}_{\text{a}}\phantom{\rule{0.2em}{0ex}}\left(\text{for}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{K}_{\text{b}}\phantom{\rule{0.2em}{0ex}}\left(\text{for}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\right)={K}_{\text{w}}=1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}$
${K}_{\text{a}}\phantom{\rule{0.2em}{0ex}}\left(\text{for}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right)\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{{K}_{\text{w}}}{{K}_{\text{b}}\phantom{\rule{0.2em}{0ex}}\left(\text{for}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\right)}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}}{4.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}}\phantom{\rule{0.2em}{0ex}}=2.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$

Now we have the ionization constant and the initial concentration of the weak acid, the information necessary to determine the equilibrium concentration of H 3 O + , and the pH: With these steps we find [H 3 O + ] = 2.3 $×$ 10 −3 M and pH = 2.64

(a) Do the calculations and show that the hydronium ion concentration for a 0.233- M solution of ${\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}$ is 2.3 $×$ 10 −3 and the pH is 2.64.

(b) What is the hydronium ion concentration in a 0.100- M solution of ammonium nitrate, NH 4 NO 3 , a salt composed of the ions ${\text{NH}}_{4}{}^{\text{+}}$ and ${\text{NO}}_{3}{}^{\text{−}}.$ Use the data in [link] to determine K b for the ammonium ion. Which is the stronger acid ${\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}$ or ${\text{NH}}_{4}{}^{\text{+}}?$

(a) ${K}_{a}\phantom{\rule{0.2em}{0ex}}\left(\text{for}\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{4}{}^{\text{+}}\right)=5.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10},$ [H 3 O + ] = 7.5 $×$ 10 −6 M ; (b) ${\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}$ is the stronger acid.

## Salts of weak acids and strong bases

When we neutralize a weak acid with a strong base, we get a salt that contains the conjugate base of the weak acid. This conjugate base is usually a weak base. For example, sodium acetate, NaCH 3 CO 2 , is a salt formed by the reaction of the weak acid acetic acid with the strong base sodium hydroxide:

${\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+\text{NaOH}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{NaCH}}_{3}{\text{CO}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(aq\right)$

A solution of this salt contains sodium ions and acetate ions. The sodium ion, as the conjugate acid of a strong base, has no effect on the acidity of the solution. However, the acetate ion, the conjugate base of acetic acid, reacts with water and increases the concentration of hydroxide ion:

${\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{OH}}^{\text{−}}\left(aq\right)$

The equilibrium equation for this reaction is the ionization constant, K b , for the base ${\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}.$ The value of K b can be calculated from the value of the ionization constant of water, K w , and K a , the ionization constant of the conjugate acid of the anion using the equation:

${K}_{\text{w}}={K}_{\text{a}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{K}_{\text{b}}$

For the acetate ion and its conjugate acid we have:

${K}_{\text{b}}\phantom{\rule{0.2em}{0ex}}\left(\text{for}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right)=\phantom{\rule{0.2em}{0ex}}\frac{{K}_{\text{w}}}{{K}_{\text{a}}\phantom{\rule{0.2em}{0ex}}\left(\text{for}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right)}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}}{1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}5.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}$

Some handbooks do not report values of K b . They only report ionization constants for acids. If we want to determine a K b value using one of these handbooks, we must look up the value of K a for the conjugate acid and convert it to a K b value.

## Equilibrium in a solution of a salt of a weak acid and a strong base

Determine the acetic acid concentration in a solution with $\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]=0.050\phantom{\rule{0.2em}{0ex}}M$ and [OH ] = 2.5 $×$ 10 −6 M at equilibrium. The reaction is:

${\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{OH}}^{\text{−}}\left(aq\right)$

## Solution

We are given two of three equilibrium concentrations and asked to find the missing concentration. If we can find the equilibrium constant for the reaction, the process is straightforward.

The acetate ion behaves as a base in this reaction; hydroxide ions are a product. We determine K b as follows:

${K}_{\text{b}}\phantom{\rule{0.2em}{0ex}}\left(\text{for}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right)\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{{K}_{\text{w}}}{{K}_{\text{a}}\phantom{\rule{0.2em}{0ex}}\left(\text{for}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right)}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}}{1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}}\phantom{\rule{0.2em}{0ex}}=5.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}$

Now find the missing concentration:

${K}_{\text{b}}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]\left[{\text{OH}}^{\text{−}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]}\phantom{\rule{0.2em}{0ex}}=5.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}$
$=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]\left(2.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\right)}{\left(0.050\right)}\phantom{\rule{0.2em}{0ex}}=5.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}$

Solving this equation we get [CH 3 CO 2 H] = 1.1 $×$ 10 −5 M .

What is the pH of a 0.083- M solution of CN ? Use 4.9 $×$ 10 −10 as K a for HCN. Hint: We will probably need to convert pOH to pH or find [H 3 O + ] using [OH ] in the final stages of this problem.

11.11

what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
how do you find theWhat are the wavelengths and energies per photon of two lines
The eyes of some reptiles are sensitive to 850 nm light. If the minimum energy to trigger the receptor at this wavelength is 3.15 x 10-14 J, what is the minimum number of 850 nm photons that must hit the receptor in order for it to be triggered?
A teaspoon of the carbohydrate sucrose contains 16 calories, what is the mass of one teaspoo of sucrose if the average number of calories for carbohydrate is 4.1 calories/g?
4. On the basis of dipole moments and/or hydrogen bonding, explain in a qualitative way the differences in the boiling points of acetone (56.2 °C) and 1-propanol (97.4 °C), which have similar molar masses
Calculate the bond order for an ion with this configuration: (?2s)2(??2s)2(?2px)2(?2py,?2pz)4(??2py,??2pz)3
Which of the following will increase the percent of HF that is converted to the fluoride ion in water? (a) addition of NaOH (b) addition of HCl (c) addition of NaF  By Rhodes     By Rhodes   