MHT CET · Chemistry · Ionic Equilibrium
What is the \(\mathrm{pH}\) of \(0.005 \mathrm{M} \mathrm{NaOH}\) solution?
- A \(2.30\)
- B \(12.6\)
- C \(11.7\)
- D \(3.2\)
Answer & Solution
Correct Answer
(C) \(11.7\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{NaOH} \longrightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-} \\ & \begin{array}{lll}0.005 \mathrm{M} & 0.005 \mathrm{M} & 0.005 \mathrm{M}\end{array} \\ & \mathrm{pOH}=-\log _{10}[\mathrm{OH}] \\ & =-\log _{10}[0.005] \\ & =-\log _{10}\left(5 \times 10^{-3}\right) \\ & =-\left[\log 10^{-3}+\log 5\right] \\ & =-[-3+0.699] \\ & =-[-2.301] \\ & =2.301 \\ & \mathrm{pOH}+\mathrm{pH}=14 \\ & \therefore \quad \mathrm{pH}=14-2.301 \\ & =11.699 \\ & \approx 11.7 \\ & \end{aligned}\)
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