A buffer solution contains 0.441 M ammonium bromide and 0.383 M ammonia. If 0.0318 moles of...
A buffer solution contains 0.441 M ammonium bromide and 0.383 M ammonia. If 0.0318 moles of potassium hydroxide are added to 250 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume does not change upon adding potassium hydroxide) pH =
Homework Answers
Given buffer contain two components. Acidic component is ammonium bromide and basic component is ammonia.
When base is added to buffer solution it reacts with acidic component and produces base. The reaction is
NH4Br + KOH 
KBr +
NH3 + H2O
[KOH ] = NO of moles / Volume of solution in L
= 0.0318 mol / 0.25 L
= 0.1272 M
Let's use ICE table
| Concentration | NH4Br | KOH | NH3 |
| Initial | 0.441 | 0.1272 | 0.383 |
| Change | -0.1272 | -0.1272 | +0.1272 |
| Equilibrium | 0.3138 | 0.0000 | 0.5102 |
pH of buffer solution is calculated by using Henderson's equation.
pH = pKa + log [ NH3 ] / [NH4Br]
= 9.24 + log 0.5102 / 0.3138
= 9.24 + 0.211
= 9.45
ANSWER :pH = 9.45
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