Question

# A 44.9 mL sample of a 0.350 M solution of NaCN is titrated by 0.260 M...

A 44.9 mL sample of a 0.350 M solution of NaCN is titrated by 0.260 M HCl. Kb 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 48.4 mL of 0.260 M HCl pH = (c) at the equivalence point pH = (d) after the addition of 76.2 mL of 0.260 M HCl. pH =

a ) Kb = 2 x 105

[OH-] = underroot Kb x c

= underroot 2 x 105 x 0.350

= 2.603 x 10 -3M

pOH = - log [OH-]

= - log 2.603 x 10 -3

= 2.58

pH = 14 - 2.58 =11.42

b)Vol after mixing = 48.4 + 44.9 = 93.3

molarity of HCL = 48.4 / 93.3 M

Molarity of NaCN = 44.9 / 93.3 M

Net molarity = 48.4 / 93.3 - 44.9 / 93.3

= 0.037

[H+] = 0.037 = 3.7 x 10-2

pH = -log  3.7 x 10-2

= 1.43

c ) At equivqlence point pH = kb = 2

d)

Vol after mixing = 76.2 + 44.9 = 121.1

molarity of HCL = 76.2 / 121.1 M

Molarity of NaCN = 44.9 / 121.1 M

Net molarity = 76.2 / 121.1 - 44.9 / 121.1

= 0.25

[H+] = 0.25 = 2.5 x 10-1

pH = -log 2.5 x 10-1

= 0.6021

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