# 15.2 Lewis acids and bases  (Page 3/12)

 Page 3 / 12

## Dissociation of a complex ion

Calculate the concentration of the silver ion in a solution that initially is 0.10 M with respect to $\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}.$

## Solution

We use the familiar path to solve this problem: 1. Determine the direction of change. The complex ion $\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}$ is in equilibrium with its components, as represented by the equation:

${\text{Ag}}^{\text{+}}\left(aq\right)+2{\text{NH}}_{3}\left(aq\right)⇌\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{\text{+}}\left(aq\right)$

We write the equilibrium as a formation reaction because Appendix K lists formation constants for complex ions. Before equilibrium, the reaction quotient is larger than the equilibrium constant [ K f = 1.7 $×$ 10 7 , and $Q=\frac{0.10}{0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0},$ it is infinitely large], so the reaction shifts to the left to reach equilibrium.

2. Determine x and equilibrium concentrations. We let the change in concentration of Ag + be x . Dissociation of 1 mol of $\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}$ gives 1 mol of Ag + and 2 mol of NH 3 , so the change in [NH 3 ] is 2 x and that of $\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}$ is – x . In summary: 3. Solve for x and the equilibrium concentrations. At equilibrium:

${K}_{\text{f}}=\frac{\left[\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}\right]}{\left[{\text{Ag}}^{+}\right]{\left[{\text{NH}}_{3}\right]}^{2}}$
$1.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\phantom{\rule{0.2em}{0ex}}\text{=}\phantom{\rule{0.2em}{0ex}}\frac{0.10-x}{\left(x\right){\left(2x\right)}^{2}}$

Both Q and K f are much larger than 1, so let us assume that the changes in concentrations needed to reach equilibrium are small. Thus 0.10 – x is approximated as 0.10:

$1.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}=\frac{0.10-x}{\left(x\right){\left(2x\right)}^{2}}$
${x}^{3}=\frac{0.10}{4\left(1.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\right)}\phantom{\rule{0.2em}{0ex}}=1.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}$
$x\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\sqrt{1.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}1.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}$

Because only 1.1% of the $\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}$ dissociates into Ag + and NH 3 , the assumption that x is small is justified.

Now we determine the equilibrium concentrations:

$\left[{\text{Ag}}^{+}\right]\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}0+x=1.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}M$
$\left[{\text{NH}}_{3}\right]=0+2x=2.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}M$
$\left[\text{Ag}{\left({\text{NH}}_{3}\right)}_{2}{}^{+}\right]=0.10-x=0.10-0.0011=0.099$

The concentration of free silver ion in the solution is 0.0011 M .

4. Check the work. The value of Q calculated using the equilibrium concentrations is equal to K f within the error associated with the significant figures in the calculation.

Calculate the silver ion concentration, [Ag + ], of a solution prepared by dissolving 1.00 g of AgNO 3 and 10.0 g of KCN in sufficient water to make 1.00 L of solution. (Hint: Because Q < K f , assume the reaction goes to completion then calculate the [Ag + ] produced by dissociation of the complex.)

2.5 $×$ 10 –22 M

## Key concepts and summary

G.N. Lewis proposed a definition for acids and bases that relies on an atom’s or molecule’s ability to accept or donate electron pairs. A Lewis acid is a species that can accept an electron pair, whereas a Lewis base has an electron pair available for donation to a Lewis acid. Complex ions are examples of Lewis acid-base adducts. In a complex ion, we have a central atom, often consisting of a transition metal cation, which acts as a Lewis acid, and several neutral molecules or ions surrounding them called ligands that act as Lewis bases. Complex ions form by sharing electron pairs to form coordinate covalent bonds. The equilibrium reaction that occurs when forming a complex ion has an equilibrium constant associated with it called a formation constant, K f . This is often referred to as a stability constant, as it represents the stability of the complex ion. Formation of complex ions in solution can have a profound effect on the solubility of a transition metal compound.

what are oxidation numbers
pls what is electrolysis
Electrolysis is the process by which ionic substances are decomposed (broken down) into simpler substances when an electric current is passed through them. ... Electricity is the flow of electrons or ions. For electrolysis to work, the compound must contain ions.
AZEEZ
thanks
Idowu
what is the basicity of an atom
basicity is the number of replaceable Hydrogen atoms in a Molecule. in H2SO4, the basicity is 2. in Hcl, the basicity is 1
Inemesit
how to solve oxidation number
mention some examples of ester
do you mean ether?
Megan
what do converging lines on a mass Spectra represent
would I do to help me know this topic ?
Bulus
oi
Amargo
what the physic?
who is albert heistein?
Bassidi
similarities between elements in the same group and period
what is the ratio of hydrogen to oxulygen in carbohydrates
bunubyyvyhinuvgtvbjnjnygtcrc
yvcrzezalakhhehuzhbshsunakakoaak
what is poh and ph
please what is the chemical configuration of sodium
Sharon
2.8.1
david
1s²2s²2p⁶3s¹
Haile
2, 6, 2, 1
Salman
1s2, 2s2, 2px2, 2py2, 2pz2, 3s1
Justice
1s2,2s2,2py2,2
Maryify
1s2,2s2,2p6,
Francis
1s2,2s2,2px2,2py2,2pz2,3s1
Nnyila
what is criteria purity
cathode is a negative ion why is it that u said is negative
cathode is a negative electrode while cation is a positive ion. cation move towards cathode plate.
king
CH3COOH +NaOH ,complete the equation
compare and contrast the electrical conductivity of HCl and CH3cooH
The must be in dissolved in water (aqueous). Electrical conductivity is measured in Siemens (s). HCl (aq) has higher conductivity, as it fully ionises (small portion of CH3COOH (aq) ionises) when dissolved in water. Thus, more free ions to carry charge.
Abdelkarim
HCl being an strong acid will fully ionize in water thus producing more mobile ions for electrical conduction than the carboxylic acid
Valentine
differiante between a weak and a strong acid
david
how can I tell when an acid is weak or Strong
Amarachi
an aqueous solution of copper sulphate was electrolysed between graphite electrodes. state what was observed at the cathode
write the equation for the reaction that took place at the anode
Bakanya
what is enthalpy of combustion
Bakanya
Enthalpy change of combustion: It is the enthalpy change when 1 mole of substance is combusted with excess oxygen under standard conditions. Elements are in their standard states. Conditions: pressure = 1 atm Temperature =25°C
Abdelkarim
Observation at Cathode: Cu metal deposit (pink/red solid).
Abdelkarim
Equation at Anode: (SO4)^2- + 4H^+ + 2e^- __> SO2 + 2H2O
Abdelkarim
Equation : CuSO4 -> Cu^2+ + SO4^2- equation at katode: 2Cu^2+ + 4e -> 2Cu equation at anode: 2H2O -> 4H+ + O2 +4e at the anode which reacts is water because SO4 ^ 2- cannot be electrolyzed in the anode
Niken By By By Madison Christian By John Gabrieli By OpenStax By Richley Crapo By Stephen Voron By Janet Forrester By Dakota Bocan By Dravida Mahadeo-J... By Abby Sharp By Michael Sag