Question

When $100$ mL of $0.1M$ $KN{O}_3$, $400$ mL of $0.2M$ $HCl$ and $500$ mL of $0.3M$ $H_{2}S{O}_4$ are mixed, then in the resulting solution :

• A. $\left[K^{\oplus }\right]=0.01M$
• B. $\left[S{O}_4^{2-}\right]=0.15M$
• C. $\left[H^{\oplus }\right]=0.38M$
• D. $\left[N{O}_3^{\ominus }\right]=0.08$ and $\left[C{l}^{\ominus }\right]=0.01M$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\left[H^{\oplus }\right]=0.38M$
B $\left[S{O}_4^{2-}\right]=0.15M$
C $\left[K^{\oplus }\right]=0.01M$

The reaction is given below :

$\underset{(100\times 0.1=10mmol)}{KN{O}_3}+\underset{(400\times 0.2=80mmol)}{HCl}+\underset{(500\times 0.3=150mmol)}{H_{2}S{O}_4}$

Total volume $=100+400+500=1000$ mL

$M_{K^{+}}=\frac{10}{1000}=0.01M$

$M_{S{O}_4^{-2}}=\frac{150}{1000}=0.15M$

$M_{H^{+}}=\frac{80+2\times 150}{1000}=0.38M$

$M_{N{O}_3^{-}}=\frac{10}{1000}=0.01M$

$M_{C{l}^{-}}=\frac{80}{1000}=0.08M$

Therefore, the options are A, B and C.