Question

A chemist titrates 100.0 ml of a 0,6612 M hydrocyanic acid (HCN) solution with 0.8512 M NaOH solution at 25 c. Calculate the pH at equivalence. The pk, of hydrocyanic acid is 9.21. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of N.OH solution added. pH- x 5 ?

Answer 1

use:

pKa = -log Ka

9.21 = -log Ka

Ka = 6.166*10^-10

find the volume of NaOH used to reach equivalence point

M(HCN)*V(HCN) =M(NaOH)*V(NaOH)

0.6612 M *100.0 mL = 0.8512M *V(NaOH)

V(NaOH) = 77.6786 mL

Given:

M(HCN) = 0.6612 M

V(HCN) = 100 mL

M(NaOH) = 0.8512 M

V(NaOH) = 77.6786 mL

mol(HCN) = M(HCN) * V(HCN)

mol(HCN) = 0.6612 M * 100 mL = 66.12 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.8512 M * 77.6786 mL = 66.12 mmol

We have:

mol(HCN) = 66.12 mmol

mol(NaOH) = 66.12 mmol

66.12 mmol of both will react to form CN- and H2O

CN- here is strong base

CN- formed = 66.12 mmol

Volume of Solution = 100 + 77.6786 = 177.6786 mL

Kb of CN- = Kw/Ka = 1*10^-14/6.166*10^-10 = 1.622*10^-5

concentration ofCN-,c = 66.12 mmol/177.6786 mL = 0.3721M

CN- dissociates as

CN- + H2O -----> HCN + OH-

0.3721 0 0

0.3721-x x x

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

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.622*10^-5)*0.3721) = 2.457*10^-3

since c is much greater than x, our assumption is correct

so, x = 2.457*10^-3 M

[OH-] = x = 2.457*10^-3 M

use:

pOH = -log [OH-]

= -log (2.457*10^-3)

= 2.6097

use:

PH = 14 - pOH

= 14 - 2.6097

= 11.3903

Answer: 11.39