MHT CET · Chemistry · Ionic Equilibrium
If \(20 \mathrm{~mL}\) of an acidic solution of \(\mathrm{pH} 3\) is diluted to \(100 \mathrm{~mL}\), the \(\mathrm{H}^{+}\) ion concentration will be
- A \(1 \times 10^{-3} \mathrm{M}\)
- B \(2 \times 10^{-3} \mathrm{M}\)
- C \(2 \times 10^{-4} \mathrm{M}\)
- D \(0.02 \times 10^{-4} \mathrm{M}\)
Answer & Solution
Correct Answer
(C) \(2 \times 10^{-4} \mathrm{M}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{pH}=3\) (given)
\(\therefore\left[\mathrm{H}^{+}\right]=1 \times 10^{-3} \mathrm{M} =1 \times 10^{-3} \mathrm{~N} \)
\( N_{1} V_{1} =N_{2} V_{2} \)
\( 1 \times 10^{-3} \times 20 =N_{2} \times 100 \)
\( N_{2} =2 \times 10^{-4} \)
\( \therefore \left[\mathrm{H}^{+}\right] =2 \times 10^{-4} \mathrm{M}\)
\(\therefore\left[\mathrm{H}^{+}\right]=1 \times 10^{-3} \mathrm{M} =1 \times 10^{-3} \mathrm{~N} \)
\( N_{1} V_{1} =N_{2} V_{2} \)
\( 1 \times 10^{-3} \times 20 =N_{2} \times 100 \)
\( N_{2} =2 \times 10^{-4} \)
\( \therefore \left[\mathrm{H}^{+}\right] =2 \times 10^{-4} \mathrm{M}\)
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