Question

A 100.0 mL sample of 0.10 M Ca(OH)2 is titrated with 0.10 M HBr. Determine the pH of the solution after the addition of 300.0 mL HBr. A) 2.62 B)2.00 C) 1.7 D) 12.52

Answer 1

Answer : The correct option is, (C) 1.7

Explanation :

First we have to calculate the moles of $Ca(OH)_2$ and $HBr$.

$\text{Moles of }Ca(OH)_2=\text{Concentration of }Ca(OH)_2\times \text{Volume of solution}=0.10M\times 0.1L=0.01mole$

$\text{Moles of }HBr=\text{Concentration of }HBr\times \text{Volume of solution}=0.10M\times 0.3L=0.03mole$

The balanced chemical reaction will be:

$Ca(OH)_2+2HBr\rightleftharpoons CaBr_2+2H_2O$

0.01 mole of $Ca(OH)_2$ dissociate to give 0.01 mole of $Ca^{2+}$ ion and 0.02 mole of $OH^-$ ion

and

0.03 mole of $HBr$ dissociate to give 0.03 mole of $H^+$ ion and 0.03 mole of $Br^-$ ion

That means,

0.02 moles of $OH^-$ ion  neutralize by 0.02 moles of $H^+$ ion.

The excess moles of $H^+$ ion = 0.03 - 0.02 = 0.01 mole

Total volume of solution = 100 + 300 = 400 ml = 0.4 L

Now we have to calculate the concentration of $H^+$ ion.

$\text{Concentration of }H^+=\frac{\text{Moles of }H^+}{\text{Total volume}}$

$\text{Concentration of }H^+=\frac{0.01mole}{0.4L}=0.025M$

Now we have to calculate the pH of the solution.

$pH=-\log [H^+]$

$pH=-\log (0.025M)$

$pH=1.7$

Therefore, the pH of the solution is, 1.7