Question

Calculate the pH of 1.6M solutions of the following salts:

(This problem requires values in your textbook's specific appendices, which you can access through the OWLv2 MindTap Reader. You should not use the OWLv2 References' Tables to answer this question as the values will not match.)

a. NaHCOO pH=?

b. KOI pH=?

c. LiCN pH=?

Answer 1

(a) It is a salt of weak acid HCOOH and strong base NaOH. The salt is basic in nature. The HCOO- hydrolyses in the aqueous solution as:

HCOO^- (aq) + H2O = HCOOH (aq) + OH^- (aq)

The dissociation constant of HCOO^- (Kb) = 10^-14/ Ka of HCOOH = 10^-14/ (1.8 * 10^-4) = 5.56 * 10^-11

[OH-] in the solution = ( [HCOO^-] * Kb)^1/2 = ( 1.6 * 5.56 * 10^-11) ^1/2 = 9.43 * 10^-6 M

pOH = - log[OH-] = - log ( 9.43 * 10^-6 ) = 5.03

pH = 14 - 5.03 = 8.97

(b)

It is a salt of weak acid HOI and strong base KOH. The salt is basic in nature. The IO- hydrolyses in the aqueous solution as:

IO^- (aq) + H2O = HOI (aq) + OH^- (aq)

The dissociation constant of IO^- (Kb) = 10^-14/ Ka of HOI = 10^-14/ (2.3 * 10^-11) = 4.35 * 10^-4

[OH-] in the solution = ( [HCOO^-] * Kb)^1/2 = ( 1.6 * 4.35 * 10^-4) ^1/2 = 0.0264 M

pOH = - log[OH-] = - log ( 0.0264 ) = 1.58

pH = 14 - 1.58 = 12.42

(c)

It is a salt of weak acid HCN and strong base LiOH. The salt is basic in nature. The CN- hydrolyses in the aqueous solution as:

CN^- (aq) + H2O = HCN(aq) + OH^- (aq)

The dissociation constant of CN^- (Kb) = 10^-14/ Ka of HCOOH = 10^-14/ (6.2 * 10^-10) =1.61* 10^-5

[OH-] in the solution = ( [HCOO^-] * Kb)^1/2 = ( 1.6 * 1.61 * 10^-5) ^1/2 = 0.0051 M

pOH = - log[OH-] = - log ( 0.0051 ) = 2.29

pH = 14 - 2.29 = 11.71