Question

Predict whether the following salt solutions are neutral, acidic, or basic, and calculate the pH of each. Hint: Write the equation for dissociation of the salt and any equilibrium established.

0.25 M NH4Br; KB of NH3 = 1.8×10−5 0.10 M NaCN;

KA of HCN = 4.9×10−10 0.20 M NaNO2;

KA of HNO2 = 4.6×10−4

Answer 1

a)

NH4Br --> NH4+ + Br-

note that NH4+ hydrolysis to form:

NH4+(aq) + H2O(l) <-> H3O+(aq) + NH3(aq)

then, H3O+ is acidic

Ka = [NH3][H3O]/[NH4+]

Ka = Kw/Kb = (10^-14)/(1.8*10^-5) = 5.55*10^-10

5.55*10^-10 = x*x/(0.25-x)

x = H = 1.17*10^5

pH = -log(1.17*10^5) = 5.07

b)

NaCN --> Na+ + CN-

CN- will hydrolyse as follow

CN- + H2O <-> HCN + OH-

OH- will make a basic solution

Kb = [HCN][OH-]/[CN-]

(10^-14)/(4.9*10^-10) = x*x/(0.2-x)

2.05*10^-5 = x*x/(0.1-x)

x = OH = 0.00142

pOH = -log(0.00142) = 2.85

pH = 14-2.85 = 11.15

c)

NaNO2 --> Na+ + NO2-

H2O + NO2- <-> HNO2 + OH-

once again, this will be basic

Kb = Kw/Ka = (10^-14)/(4.6*10^-4) = 2.17*10^-11

2.17*10^-11 = x*x/(0.2-x)

x = OH = 2.083*10^-6

pOH = -log(2.083*10^-6) = 5.68131

pH = !4-5.68131

pH = 8.31869