Question

A 48.0 ml volume of 0.25 m hbr is titrated with 0.50 m koh. calculate the ph after addition of 24.0 ml of koh at 25 ∘c. express the ph numerically.

Answer 1

Answer: pH=7

Explanation:

Moles can be calculated by using the formula:

$\text{no of moles of HBr}={\text{Molarity}\times {\text{Volume in L}}$

$\text{no of moles}={0.25M}\times {0.048L}=0.012$

1) For 24 ml of KOH added

$\text{no of moles of KOH}={\text{Molarity}\times {\text{Volume in L}}$

$\text{no of moles}={0.50M}\times {0.024L}=0.012$

$HBr+KOH\rightarrow KBr+H_2O$

According to given balanced equation, 1 mole of KOH neutralize 1 mole of HBr

Thus 0.012 moles of KOH will neutralize 0.012 moles of HBr and the solution becomes neutral, thus the pH will be 7.