Question

Find the pH of resultant solution when $100ml$ of $0.005M$ $H_2{SO}_4$ at ${25}^{o}C$ is diluted to $1000ml$. What is the amount of $NaOH$ required to be dissolved in $500ml$ to exactly neutralize the above solution?

• A. $4g$
• B. $4\times {10}^{-2}g$
• C. $40g$
• D. $0.4g$

Answer 1

The correct option is B $4\times {10}^{-2}g$

Number of moles of $H^{+}$ in $0.005M{H}_{2}S{O}_4=0.005\times 2\times 0.1=0.001moles$

After dilution concentration of $H^{+}$ ions= $\frac{0.001}{1}=0.001mole/l$

$pH=-log\left[0.001\right]=3$

Amount of $NaOH$ required to neutralize the solution= $0.001mole$

Weight= mole $\times$ molecular weight $=0.001\times 40=4\times {10}^{-2}g$