Question

A 46.3 mL sample of a 0.380 M solution of NaCN is titrated by 0.350 M HCl. K_b for CN^- is 2.0×10^-5. Calculate the pH of the solution:

(a) prior to the start of the titration	pH =
(b) after the addition of 25.1 mL of 0.350 M HCl	pH =
(c) at the equivalence point	pH =
(d) after the addition of 71.9 mL of 0.350 M HCl.	pH =

Answer 1

for part (a)

pH prior to start of titration

@ 0.380 M NaCN and 0.350 M HCl

pH = 3.181 and pOH = 10.819

for part (b)

we know

  

so

from the above table the equation becomes,

on substituting the values we get x = 4.08 E-6

so pH = 5.38

for part (c) as equivalance point reaches in part (b) the answer for pH is same as 5.38

for part (d)

adding 71.9 ml is beyond equivalence point so the pH depends on excess strong acid

[H+] = {(0.0719)(0.1) - (0.05)(0.05)}/(0.0719 + 0.05)

       = 0.03847

pH = 1.414 strong acidic