Question

If $100$ mL of $1.0M$ $H_{2}S{O}_4$ is mixed with $200$ mL of $1.0M$ $KOH$ solution, molarity of the salt formed is :

• A. $0.100M$
• B. $0.200M$
• C. $0.333M$
• D. $0.666M$

Answer 1

Answer: (C)

$0.333M$

The neutralization reaction is as shown below.
$H_{2}S{O}_4+2KOH\to K_{2}S{O}_4+2H_{2}O$
$0.1$ mol of sulfuric acid will react with $0.2$ mol of $KOH$ to form $0.1$ mol of potassium sulphate.
Total volume is $300$ mL.
The molarity of the salt formed is $\left[K_{2}S{O}_4\right]=\frac{0.1}{0.3}=0.333M$