Question

The pKa of hypochlorous acid is 7.530. A 53.0 mL solution of 0.100 M sodium hypochlorite (NaOCl) is titrated with 0.303 M HCl. Calculate the pH of the solution

a) after the addition of 6.47 mL of 0.303 M HCl.

b) after the addition of 18.4 mL of 0.303 M HCl.

c) at the equivalence point with 0.303 M HCl.

Answer 1

a) after the addition of 6.47 mL of 0.303 M HCl.

Solution :- lets first calculate the moles of the NaOCl and HCl

Moles = molarity *volume in liter

Moles of NaOCl = 0.100 mol per L * 0.053 L =0.0053 mol

Moles of HCl =0.303 mol per L *0.00647L=0.00196 mol

Moles of HCl are less than moles of NaOCl

Therefore after reaction moles of NaOCl remain = 0.0053 – 0.00196 = 0.00334 mol

And moles of HOCl formed = 0.00196 mol

Now lets calculate new molarity of both at total volume

Total volume = 53.0 ml + 6.47 ml = 59.47 ml =0.05947 L

New molarity of the NaOCl =0.00334 mol / 0.05947 L =0.05616 M

New molarity of the HOCl = 0.00196 mol / 0.05947 L =0.03296 M

Now using the Henderson equation we can calculate the pH

pH =pka + log ([base]/[acid])

pH = 7.53+log([0.05616]/[0.03296])

pH = 7.76

b) after the addition of 18.4 mL of 0.303 M HCl.

Solution :-

Moles of NaOCl = 0.100 mol per L * 0.053 L =0.0053 mol

Moles of HCl =0.303 mol per L *0.0184 L=0.005575 mol

Moles of HCl are more than moles of NaOCl

Therefore after reaction all the NaOCl converted into HOCl

Therefore the excess moles of HCl remain are = 0.005575 – 0.0053 = 0.000275 mol HCl

Now lets calculate new molarity of HCl

Total volume = 53.0 ml + 18.4 ml = 71.4 ml =0.0714 L

New molarity of the HCl = 0.000275 mol / 0.0714 L = 0.00385 M

HCl is strong acid therefore it will dissociate completely so lets calculate the pH using the concentration of the HCl

pH= -log[H+]

pH= -log[0.00385]

pH= 2.41

c) at the equivalence point with 0.303 M HCl.

Solution

At the equivalence point moles of acid and moles of base are same

Therefore moles of HCl = 0.0053

Lets calculate the volume of the HCl

Volume = 0.0053 mol / 0.303 mol per L = 0.0175 L = 17.5 ml

All the moles of OCl- will convert to HOCl

So the moles of HOCl formed = 0.0053

And total volume = 53.0 ml + 17.5 ml = 70.5 ml = 0.0705 L

Now lets calculate the molarity of the HOCl

[HOCl] =0.0053 mol /0.0705 L =0.07518 M

We have pka of the HOCl using that lets find ka of HOCl

Pka = -log ka

Ka = antilog [-pka]

Ka = antilog [-7.530]

Ka = 2.95*10^-8

Now lets calculate the H3O+ using the ka value

HOCl + H2O ------ > H3O+ + OCl-

0.07518                        0               0

-x                                  +x             +x

0.07518-x                      x               x

Ka = [H3O+][OCl-]/[HOCl]

2.95*10^-8= [x][x]/[0.07518-x]

Since ka value is very small therefore neglect x from denominator then we get

2.95*10^-8= [x][x]/[0.07518]

2.95*10^-8 *0.07518= x²

2.218*10^-9 =x²

By taking square root of both side we get

4.71*10^-5 =x

So lets calculate pH using this concentration

pH= -log [H3O+]

pH= -log[4.71*10^-5]

pH = 4.33