Question

$50$ml of $0.1$M. HCl is titrated with $0.1$ M.NaOH. At pH = $3$. vol of NaOH used is (approximately) :

• A. $49$ml
• B. $50$ml
• C. $45$ml
• D. $41$ml

Answer 1

Answer: (A)

$49$ml

Suppose V mL of 0.1 M NaOH is used.

Number of moles of NaOH $=\frac{VmL}{1000mL/L}\times 0.1mol/L={10}^{-4}Vmol$

Number of moles of HCl $=\frac{50mL}{1000mL/L}\times 0.1mol/L=0.005mol$

Out of 0.005 moles of HCl, ${10}^{-4}V$ moles will be neutralised with ${10}^{-4}V$ moles of NaOH.

$0.005-{10}^{-4}V$ moles of HCl will remain.

$0.005-{10}^{-4}V$ moles of HCl will give $0.005-{10}^{-4}V$ moles of $H^{+}$ ions.

Total volume $=(50+V)mL=\frac{(50+V)mL}{1000mL/L}=(0.05+0.001V)L$
$\left[H^{+}\right]=\frac{(0.005-{10}^{-4}V)mol}{(0.05+0.001V)L}$......(1)
$pH=3$

$\left[H^{+}\right]={10}^{-pH}={10}^{-3}M$......(2)

But, (1) $=$ (2)
$\frac{(0.005-{10}^{-4}V)mol}{(0.05+0.001V)L}={10}^{-3}M$

$\frac{(0.005-{10}^{-4}V)}{(0.05+0.001V)}={10}^{-3}$......(3)

Assuming V to approximately equal to 50, we can assume that the denominator $0.05+0.001V\simeq 0.05+0.001\left(50\right)=0.05+0.05=0.1$

Hence, equation (3) becomes

$\frac{(0.005-{10}^{-4}V)}{0.1}={10}^{-3}$

$0.005-{10}^{-4}V={10}^{-4}$
$0.005={10}^{-4}(1+V)$
$1+V=50$
$V=49mL$
Hence, 49 mL of 0.1 M NaOH is used.