Question

A 100.0-mL buffer solution is 0.175 M in HClO and 0.150 M in NaClO.

A/ What is the initial pH of this solution?

B/What is the pH after addition of 150.0 mg of HBr?

C/What is the pH after addition of 85.0 mg of NaOH?

Answer 1

A)

pka for = 7.53

For buffers,

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

pH = 7.53 + log [ 0.150 / 0.175]

pH = 7.46

B)

Number of moles of HBr = mass /molar mass

= 150 x 10^-3 g / 80.91 gmol-1

= 1.85 x 10^-3 mol

Similarly, Mol of HClO = 0.175 x 0.1 L

= 0.0175 mol

Total mol of acid = (1.85 x 10^-3 + 0.0175) mol = 0.01935 mol

Mol of Base = 0.15 M x 0.1 L = 0.015 mol

Mol of base after acid addition = ( 0.015 - 1.85 x 10^-3) mol

= 0.01315 mol

So, pH = 7.53 + log( ( 0.01315 /0.1 L ) / (0.01935 / 0.1 L))

pH = 7.36

C)

Mol of NaOH = 0.085 g / 40 gmol-1

= 0.0021 mol

Mol of base after addition of NaOH = 0.015 + 0.0021

= 0.0171 mol

Mol of acid after base addition = 0.0175 - 0.0021

=0.0154

So, pH = 7.53 + log ( ( 0.0171 mol / 0.1 L) / (0.0154 mol / 0.1 L))

pH = 7.57