Question

1.00 mL of 0.100 M HCl are added to 40.00 mL of buffer and the pH changes from 7.203 to 7.145. What is the buffer capacity of this buffer against acid? (Express your answer with 3 sig figs)

complete work please

Answer 1

Moles of HCl added = (molarity HCl) * (volume HCl)

Moles of HCl added = (0.100 M) * (1.00 mL)

Moles of HCl added = 0.100 mmol

Since HCl is a strong acid,

moles H⁺ = moles HCl added

moles H⁺ = 0.100 mmol

Total volume = (volume HCl) + (volume buffer)

Total volume = (1.00 mL) + (40.00 mL)

Total volume = 41.00 mL

Concentration of acid = (moles H⁺) / (Total volume)

Concentration of acid = (0.100 mmol) / (41.00 mL)

Concentration of acid = 2.44 x 10^-3 M

Change in pH = initial pH - final pH

Change in pH = (7.203) - (7.145)

Change in pH = 0.058

Buffer capacity = (Concentration of acid) / (Change in pH)

Buffer capacity = (2.44 x 10^-3 M) / (0.058)

Buffer capacity = 0.0421 M/pH