ExamBro
ExamBro
MHT CET · Chemistry · Ionic Equilibrium

The pH of \(0.005 \mathrm{M} \mathrm{KOH}\) is 9.95. Calculate the \(\left[\mathrm{OH}^{-}\right]\)

  1. A \(6.71 \times 10^{-4} \mathrm{M}\)
  2. B \(1.12 \times 10^{-4} \mathrm{M}\)
  3. C \(4.45 \times 10^{-5} \mathrm{M}\)
  4. D \(8.91 \times 10^{-5} \mathrm{M}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(8.91 \times 10^{-5} \mathrm{M}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \mathrm{P}^{\mathrm{H}}+\mathrm{P}^{\mathrm{OH}}=14 \\ & \mathrm{P}^{\mathrm{OH}}=14-.9 .95 \\ & =4.05 \\ & \mathrm{P}^{\mathrm{OH}}=-\log _{10}\left[\mathrm{OH}^{-}\right] \\ & {\left[\mathrm{OH}^{-}\right]=10^{-\mathrm{P}^{\mathrm{OH}}}=10^{-4.05}} \\ & =10^{-5+0.95}=10^{-5} \times 10^{0.95}=8.91 \times 10^{-5} \mathrm{M}\end{aligned}\)