KCET · Chemistry · Ionic Equilibrium
\(0.023 \mathrm{~g}\) of sodium metal is reacted with \(100 \mathrm{~cm}^{3}\) of water. The \(\mathrm{pH}\) of the resulting solution is
- A 10
- B 8
- C 9
- D 12
Answer & Solution
Correct Answer
(D) 12
Step-by-step Solution
Detailed explanation
\(2 \mathrm{Na}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+\mathrm{H}_{2}\)
\((2 \times 23=46) \quad(2 \times 40=80)\)
\(0.023 \quad 0.04\)
\({\left[\mathrm{OH}^{-}\right]=\frac{0.04}{40} \times 10=10^{-2} }\)
\(\because \quad \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]\)
\(\mathrm{pOH}=2\)
\(\because \quad \mathrm{pH}+\mathrm{pOH}=14\)
\(\mathrm{pH}=14-\mathrm{pOH}\)
\(\mathrm{pH}=14-2\)
\(\mathrm{pH}=12\)
\((2 \times 23=46) \quad(2 \times 40=80)\)
\(0.023 \quad 0.04\)
\({\left[\mathrm{OH}^{-}\right]=\frac{0.04}{40} \times 10=10^{-2} }\)
\(\because \quad \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]\)
\(\mathrm{pOH}=2\)
\(\because \quad \mathrm{pH}+\mathrm{pOH}=14\)
\(\mathrm{pH}=14-\mathrm{pOH}\)
\(\mathrm{pH}=14-2\)
\(\mathrm{pH}=12\)
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