Question

Question #c1b5b

Answer 1

The pH of a buffer solution depends on the ratio of concentrations of an acid and its conjugate base, using the Henderson-Hasselbalch equation.
`pH=pK_a-log_10(([HA])/([A^-]))`

Explanation:

In this example, the acid is `NH_4^+`, which has a `pK_a=9.24`

Total volume of solution = 30mL + 15 mL = 45 mL = 0.45 L

Acid concentration: `(0.1M times 0.015L)/(0.045L) = 0.033M`

Base concentration: `(0.2M times 0.030L)/(0.045L) = 0.133M`

`pH=pK_a-log_10(([HA])/([A^-]))=9.24-log_10(0.033/0.133)=9.84`

Sanity Check:
If the acid and base concentrations were equal, the `pH` would be equal to the `pK_a`, but here the base concentration is 4X the acid concentration, so the solution is a little bit more basic and the `pH` is a little bit higher than `pK_a`.