Problem: Calculate the pH of a buffer that is 0.058 M HF and 0.058 M LiF. The Ka for HF is 3.5 x 10-4. A) 9.31 B) 2.86 C) 4.69 D) 3.46 E) 10.54

Henderson-Hasselbalch equation.

$\boxed{\mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{a}}\mathbf{+}\mathbf{log}\mathbf{ }\mathbf{\left(}\frac{\mathbf{conjugate}\mathbf{ }\mathbf{base}}{\mathbf{weak}\mathbf{ }\mathbf{acid}}\mathbf{\right)}}$

Given solution:

• HF is a weak acid
• the anion in KF is the conjugate base

            ▪ F^- → anion → conjugate base

                        - concentration of F^- = concentration of KF

• The solution is composed of a weak acid and its conjugate base, therefore we have a buffer.

We can use the Henderson-Hasselbalch Equation to solve for pH.

Recall:

${\mathbf{pK}}_{\mathbf{a}}\mathbf{=}\mathbf{-}\mathbf{log}\mathbf{\hspace{0.17em}}{\mathbf{K}}_{\mathbf{a}}$

Substitute in the Henderson-Hasselbalch equation:

$\boxed{\mathbf{pH}\mathbf{=}\mathbf{-}\mathbf{log}\mathbf{ }{\mathbf{K}}_{\mathbf{a}}\mathbf{+}\mathbf{log}\mathbf{ }\mathbf{\left(}\frac{\mathbf{conjugate}\mathbf{ }\mathbf{base}}{\mathbf{weak}\mathbf{ }\mathbf{acid}}\mathbf{\right)}}$