MHT CET · Chemistry · Ionic Equilibrium
The \(\mathrm{pH}\) of a \(0.1 \mathrm{M}\) solution of \(\mathrm{NH}_{4} \mathrm{OH}\) (having \(K_{b}=1.0 \times 10^{-5}\) ) is equal to
- A 10
- B 6
- C 11
- D 12
Answer & Solution
Correct Answer
(C) 11
Step-by-step Solution
Detailed explanation
\(\left[\mathrm{OH}^{-}\right] =\sqrt{K_{b} \times C} \)
\( =\sqrt{1 \times 10^{-5} \times 10^{-1}}=\sqrt{10^{-6}}=10^{-3} \)
\( K_{w} =\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \)
\( 10^{-14} =\left[\mathrm{H}^{+}\right]\left[10^{-3}\right] \)
\(\left[\mathrm{H}^{+}\right] =10^{-11} \)
\( \text { Hence, } \mathrm{pH} =-\log \mathrm{H}^{+} \)
\( =-\log \left(1 \times 10^{-11}\right)=11\)
\( =\sqrt{1 \times 10^{-5} \times 10^{-1}}=\sqrt{10^{-6}}=10^{-3} \)
\( K_{w} =\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \)
\( 10^{-14} =\left[\mathrm{H}^{+}\right]\left[10^{-3}\right] \)
\(\left[\mathrm{H}^{+}\right] =10^{-11} \)
\( \text { Hence, } \mathrm{pH} =-\log \mathrm{H}^{+} \)
\( =-\log \left(1 \times 10^{-11}\right)=11\)
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