Question

Using tabulated Ka and Kb values, calculate the pH of the following solutions.

Answer to 2 decimal places.

A solution containing 1.9×10-1 M HNO2 and 2.4×10-1 M NaNO2.

A solution made by mixing 110.00 mL of 1.100 M CH₃COOH with 35.00 mL of 0.750 M NaOH.

This is the second time I'm posting this question so I hope someone can explain it clearly.

Answer 1

a) A solution containing 1.9×10-1 M HNO2 and 2.4×10-1 M NaNO2.

It will form a buffer

As per Hendersen's equation, we can calcualte the ph of buffer as

pH = pKa + log[salt] /[acid]

[salt]= [NaNO2] = 0.24 M

[acid] = [HNO2] = 0.19 M

pH = 3.39 + log 0.24 / 0.19 = 3.39+0.101 = 3.491

b) A solution made by mixing 110.00 mL of 1.100 M CH₃COOH with 35.00 mL of 0.750 M NaOH.

Moles of acetic acid = volume X concentration = 0.110 L X 1.1 = 0.121 moles

Moles of NaOH added = Volume X concentration = 0.035 L X 0.75 = 0.02625 moles

The moles of salt formed due to reaction of base and acid as

CH3COOH + NaOH --> CH3COO-Na+

will be = 0.02625 moles

Moles of acid left = moles of acid initial - moles of base added = 0.121 - 0.02625 = 0.09475 moles

Now again we are left with a mixture of salt and weak acid so it will act as buffer

the pH of buffer will be

pH = pKa + log [salt] / [acid]

pKa of acetic acid = 4.75

pH = 4.75 + log [0.02625 / 0.09475] = 4.75- 0.557 = 4.193