Question

A 75.0-mL sample of 0.0500 M HCN (Ka = 6.2e-10) is titrated with 0.212 M NaOH. What is the [H+] on the solution after 3.0 mL of 0.212 M NaOH have been added?

I know the answer is 3.0 x 10^-9 M but I am not sure about the steps please help

Answer 1

Given:

[HCN]=0.05 M, volume = 75.0 mL,

[NaOH]= 0.212 M, volume = 3.0 mL

Solution:

Write the balanced reaction between HCN and NaOH

HCN (aq) + NaOH (aq) --- > NaCN(aq) + H2O (l)

Write net ionic equation.

HCN (aq) + OH^- (aq) --- > CN^- (aq) + H2O (l)

From the reaction we can see that acid HCN forms its conjugate base CN^- .

When there is such conjugate acid base pair present then we can use Henderson equation to find its pH.

Henderson Hasselbalch equation.

pH = pka + log ([conj. Base ] /[weak acid])

pka = - log ka = - log (6.20E-10) = 9.20

here conjugate base is CN^- and its concentration = concentration of NaCN.

Weak acid is HCN

Calculation of equilibrium moles and concentration.

Mol HCN = Volume in L x molarity = 0.075 L x 0.05 M = 0.00375 mol

Mol NaOH = 0.003 L x 0.212 M = 0.000636 mol.

Lets find out equilibrium moles and then concentration .

HCN (aq)         +            OH^- (aq) --- >     CN^- (aq) + H2O (l)

I           0.00375                          0.000636            0                

E        (0.00375-0.000636)               0              0.000636            

We now can calculate

[HCN]= 0.00311 / total volume = 0.00311 / (0.003+0.075) = 0.03992 M

[CN-]= 0.000636 mol/ (0.003+0.075) = 0.008154 M

Plug in the values in Henderson Hasselbalch equation

pH = 9.20 + log ( 0.008154 /0.03992)

pH = 8.52

Calculation [H₃O⁺ ] by using pH formula

pH = - log [H₃O⁺]

Rewrite this formula in terms of [H₃O⁺]

[H₃O⁺]= Antilog (-pH)

[H₃O⁺] = 3.035 E-9 M

Which is equal to the 3.0 x 10^-9 M