TS EAMCET · Chemistry · Ionic Equilibrium
\(75 \mathrm{~mL}\) of \(0.2 \mathrm{M} \mathrm{HCl}\) is mixed with \(25 \mathrm{~mL}\) of \(1 \mathrm{M}\) \(\mathrm{HCl}\). To this solution, \(300 \mathrm{~mL}\) of distilled water is added. What is the \(\mathrm{pH}\) of the resultant solution?
- A \(1\)
- B \(2\)
- C \(4\)
- D \(0.2\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
\begin{aligned} V_1 & =75 \mathrm{~mL}, V_2=25 \mathrm{~mL} \\ M_1 & =0.2 \mathrm{M}, M_2=1 \mathrm{M} \\ V_3 & =300 \mathrm{~mL}, \mathrm{pH}=? \\ V & =V_1+V_2=75+25=100 \mathrm{~mL} \\ M & =\frac{V_1 M_1+V_2 M_2}{V_1+V_2}=\frac{75 \times 0.2 \times 25+1}{75+25}=0.4 \mathrm{M}…
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