Question

The $pH$ of a solution obtained by mixing of $100\, mL$ of a $HCl$ solution of $pH = 2$ with $400\, mL$ of another $HCl$ solution of $pH = 3$ will be

• A. 2
• B. 3
• C. 2.5
• D. 2.8

Answer 1

The Correct Option is C

The correct answer is option (C): 2.5
\(100\, mL\) of a \(HCl\) solution of \(p H=2\)
\(\Rightarrow\left[H^{+}\right]=10^{-2} 400\, mL\) of a \(HCl\) 
solution of \(pH=3\) \(\Rightarrow\left[H^{+}\right]=10^{-3}\) 
Resulting \(\left[H^{+}\right]=\frac{100 \times 10^{-2}+400 \times 10^{-3}}{500}\) \(=\frac{1+0.4}{500}\)
\(=\frac{1.4}{500}=0.0028\)
\(\because p H=-\log \left[H^{+}\right] \therefore p H=-\log (0.0028)\)
\(=2.5\)