Suggested Problems for Quiz on 31 Jan
15.101 For each of the following reactions, identify the Lewis acid and the Lewis base.
a. 2 Cl- + BeCl2  BeCl42b. Mg2+ + 6 H2O  Mg(H2O)62+
16.1 b. Write a balanced net ionic equation for the neutralization of equal molar amounts
of HBr and NH3. Indicate whether the pH after neutralization is greater than, equal to, or
less than 7.
16.2 c. Write a balanced net ionic equation for the neutralization of HF and NH3, and
calculate the value of Kn for the neutralization reaction.
16.3 Calculate the concentrations of all species present, the pH, and the percent
dissociation of HCN (Ka = 4.9 * 10-10) in a solution that is 0.025 M in HCN and 0.010 M
in NaCN.
16.48 Phenol (C6H5OH, Ka = 1.3 * 10-10) is a weak acid used in mouthwashes, and
pyridine (C5H5N, Kb = 1.8 * 10-9) is a weak base used as a solvent. Calculate the value of
Kn for neutralization of phenol by pyridine. Does the neutralization reaction proceed
very far toward completion?
16.50 Which of the following substances affects the percent dissociation of HNO2?
a. NaNO2
b. NaCl
c. HCl
d. Ba(NO2)2
16.52 Does the pH increase, decrease, or remain the same on addition of each of the
following?
a. LiF to an HF solution
b. KI to an HI solution
c. NH4Cl to an NH3 solution
15.101 a. Cl- is the electron donor and is therefore the Lewis base. BeCl2 accepts the
electrons and is therefore the Lewis acid.
b. Mg2+ is the electron-deficient metal cation and is therefore the Lewis acid.
H2O donates the electrons and is the Lewis base.
16.1 b. HBr is a strong enough acid that it will completely dissociate. NH3, on the other
hand, is a weak base.
Therefore, we will have a strong acid-weak base situation:
H3O+(aq) + NH3(aq)   H2O(l) + NH4+(aq)
After neutralization, water (neutral) and a potential proton donor (acid)
are present, so the pH should be below 7.
16.2 c. HF(aq) + NH3(aq)   F-(aq) + NH4+(aq)
(this is a weak acid/weak base equilibrium)
Kn = KaKb(1/Kw) = (3.5 * 10-4)(1.8 * 10-5)/(1/1.0 * 10-14) = 6.3 * 105
HCN + H2O   H3O+ + CN----------------------------------------------------------------------I(M)
0.025
~0
0.010
C(M)
-x
+x
+x
E(M)
0.025-x
x
0.010+x
16.3
Ka = 4.9 * 10-10 = [H3O+][CN-]/[HCN] = (x)(0.010+x)/0.025-x
x is small enough, so = x(0.010)/0.025
Solving for x… 1.225 * 10-11 = 0.010 + x
x = 1.23 * 10-9 = [H3O+]
pH = -log (1.23 * 10-9) = 8.91
As for the concentrations of everything else…
[HCN] = 0.025 – x = 0.025 M
[CN-] = 0.010 – x = 0.010 M
[OH-] = Kw/[H3O+] = 1.0 * 10-14/1.23 * 10-9 = 8.2 * 10-6 M
Percent dissociation = [HCN] diss/[HCN] original * 100%
= 1.23 * 10-9 M / 0.025 M * 100% = 4.9 * 10-6 %
(not very dissociated!)
16.48 We are told that both the acid and the base are weak, so we can calculate Kn using
the equation
Kn = KaKb(1/Kw) = (1.3 * 10-10)(1.8 * 10-9)(1/1 * 10-14) = 2.3 * 10-5
Since the Kn value is small, it is assumed the reaction doesn’t proceed very far
toward completion.
16.50 Look at the reaction, and think of whether each is a source of reactants or
products:
HNO2(aq) + H2O(l)   NO2-(aq) + H3O+(aq)
a. will shift the reaction to the reactant side, % dissoc decreases
b. won’t affect anything
c. provides a source of H+, contributing to H3O+, shifting reaction to the reactant
side, % dissoc decreases
d. will again shift the reaction to the reactant side, % dissoc decreases
16.52 a. If you add F-, the conjugate base, shifts the equilibrium toward reactants – less
acid dissociated, less [H3O+], higher pH.
b. HI is a strong acid and will completely dissociate. Therefore, the reaction
equilibrium being dealt with involves the dissociation of water. Addition of KI does
nothing to the equilibrium, and the pH does not change.
c. NH4Cl will add NH4+ to the reaction mixture. Since NH3 is a base, NH4+ is a
reaction product. The reaction will shift to the reactant side, and the amount of OH- in
the reaction mixture decreases, as will the pH.