Question

Calculate the pH of solution obtained by mixing 10 mL of 0.1 M HCl and 40 mL of 0.2 M $H_{2}S{O}_4.$ Take log 3.4 = 0.53

• A. 0.34
• B. 0.2
• C. 0.47
• D. None of the above

Answer 1

The correct option is C 0.47

Given is the case of a mixture of 2 strong acids.
$\text{Millimoles of}{H}^{+}\text{from HCl}=10\times 0.1=1$
Millimoles of $H^{+}$ from $H_{2}S{O}_4=40\times 0.2\times 2=16$
$\therefore \text{Total millimoles of}{H}^{+}\text{in solution}=1+16=17$
$\because [H^{+}{]}_1=\frac{\text{total Millimoles}}{V_{t}inmL}$
$\therefore \left[H^{+}\right]=\frac{17}{50}=3.4\times {10}^{-1}$
$\therefore pH=-log\left[H^{+}\right]=-log0.34$
$pH=0.47$