Question

consider the titration of a 20ml sample of .105M HCN with .125M NaOH(ka=4.9*10^-10 of HCN)

Draw a scheme of how the titration curve should look like

Find:

initial pH, pH when 10ml of NaOH were added, pH at the equivalence point, pH when 16.8ml of NaOH were added

Answer 1

(a)

Initial pH

HCN      H⁺ + CN^-

Ka of HCN = 4.9 x 10^-10

Ka = [H+] [CN-] / [HCN]

4.9 x 10^-10 = (x) (x) / (0.105)

x² = 5.1 x 10^-11

x = 7.1 x 10^-6

So, [H+] = x = 7.1 x 10^-6

pH = - log [H+] = - log (7.1 x 10^-6) = 5.15

(b)

10 mL of 0.125 M NaOH added

Now,

moles of HCN = 0.02 L x 0.105 M = 0.0021 moles

moles of NaOH = 0.01 L x 0.125 M = 0.00125 moles

When 0.00125 moles NaOH will add, it will react with 0.00125 moles HCN.

So, the excess moles of HCN left = 0.0021 moles - 0.00125 moles = 0.00085 moles

Total volume = 20 mL + 10 mL = 30 mL = 0.03 L

[HCN] = 0.00085 moles / 0.03 L = 0.028 M

HCN      H⁺   + CN^-

Ka of HCN = 4.9 x 10^-10

Ka = [H+] [CN-] / [HCN]

4.9 x 10^-10 = (x) (x) / (0.028)

x² = 1.4 x 10^-11

x = 3.74 x 10^-6

So, [H+] = x = 3.74 x 10^-6

pH = - log [H+] = - log (3.74 x 10^-6) = 5.43

(c)

At the equivalence point

At the equivalence point, moles of acid = moles of base = 0.0021 moles.

So, 0.0021 moles of NaOH will react with 0.0021 moles of HCN. So, there will be no HCN but only left with CN^-.

So, moles of CN^-= 0.0021 moles

Now,

Molarity = Moles / Liter

Liter = Moles / Molarity

Volume required for 0.0021 moles of NaOH to form 0.125 M = 0.0021 moles / 0.125 M

= 0.0168 L

= 16.8 mL

Total volume = 20 mL + 16.8 mL = 36.8 mL = 0.0368 L

So, [CN-] = 0.0021 mole / 0.0368 L = 0.06 M

Now,

CN^- + H₂O     OH^- + HCN

Ka of HCN = 4.9 x 10^-10

Kb of HCN = (1.0 x 10^-14) / (4.9 x 10^-10) = 2.0 x 10^-5

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

2.0 x 10^-5 = (x) (x) / 0.06

x² = 1.2 x 10^-6

x = 1.1 x 10^-3

So, [OH-] =  x = 1.1 x 10^-3

pOH = -log [OH-] = - log (1.1 x 10^-3) = 2.96

pH = 14 - pOH = 14 - 2.96 = 11.04

(d) pH when 16.8ml of NaOH were added

Its is same as equivalence point (answer of c).