A chemist starts with a 25.00 mL of 0.125 M NaCN solution. Please answer all the

Question

A chemist starts with a 25.00 mL of 0.125 M NaCN solution. Please answer all the following questions with reference to this original solution. K_A (HCN) = 6.31 times 10^-10. a) What is the pH of this solution (to two decimal places)? b) What volume of 0.100 M HNO_3 is required to adjust the pH of the original solution to 9.00? c) What volume (in mL) of 0.100 M HNO_3 is required to titrate the original solution to the equivalence point? Also predict whether the pH at that point will be acidic, basic or neutral.

Explanation / Answer

A)

this is a salt, NaCN, let CN- be A-

A- is formed

the next equilibrium is formed, the conjugate acid and water

A- + H2O <-> HA + OH-

Kb by definition since it is an base:

Kb = [HA ][OH-]/[A-]

Ka can be calcualted as follows:

Kb = Kw/Ka = (10^-14)/(6.31*10^-10) = 1.58*10^-5

and;

assume x = [OH-]

[OH-] = x= [HA] due to equilibrium

[A-] = M-x = 0.125-x

Kb = [HA ][OH-]/[A-]

1.58*10^-5 = x*x/(0.125-x)

solve for x with quadratic equation

x = OH- =0.001397

[OH-]  =0.001397 M

pOH = -log(OH-) = -log(0.001397) = 2.854

pH = 14-2.854= 11.4

pH = 11.4

b)

what Volume of M = 0.1 HNO3 is rquired to adjust to pH = 9

a buffer will form:

pH = pKa + log(NaCN/HCN)

pKa = -log(6.31*10^-10) = 9.19997

9 = 9.19997 + log(NaCN / HCN)

10^(9-9.199) = NaCN / HCN

0.6324*HCN = NaCN

note that, initially

mmol of NaCN = 25*0.125 = 3.125 mmol

mmol of HCN = 0

after adding mmol of HNO3 = M*V = 0.1*Vbase

mmol of NaCN = 3.125 - 0.1*Vbase

mmol of HCN = 0 + 0.1*Vbase

substittue

0.6324*HCN = NaCN

0.6324*(0.1*Vbase)= 3.125 - 0.1*Vbase

0.06324*Vbase +0.1*Vbase = 3.125

vbase = 3.125 /0.16324 = 19.14 mL approx

c)

find volume required for titration

mmol of NaCN = MV = 25*0.125 = 3.125 mmol of NaCN

so we require 3.125 mmol of H+

M = mol/V

V = mol/M = 3.125 /(0.1) = 31.25 mL of HNO3 required

the pH in equivalence:

will be acidic, since HCN is formed, so it donates patially H+ ions to solution

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