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20 mL of 0.1 M \mathrm{NH_{4}OH} is mixed with 40 mL of 0.05 M \mathrm{HCl}. The pH of the mixture is nearest to :

\\(\text{Given} : \mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=1 \times 10^{-5}, \log 2=0.30, \log 3=0.48, \log 5=0.69, \log 7=0.84, \log 11= 1.04)

Option: 1

3.2
 


Option: 2

4.2


Option: 3

5.2

 


Option: 4

6.2


Answers (1)

best_answer

\mathrm{NH_{4}OH} is weak base and \mathrm{HCl} is strong acid.

Salt \mathrm{NH_{4}OH} is forming with \mathrm{H_{2}O}.

We know,

\mathrm{\begin{aligned} &\mathrm{pH}=7-\frac{P_{K_{b}}}{2}-\frac{1}{2} \log C \\ &\mathrm{pH}=7-\frac{1}{2} \times 5 \quad-\frac{1}{2} \log \left[\mathrm{NH}_{4} \mathrm{Cl}\right] \end{aligned}}

Now,

\mathrm{\left[\mathrm{NH}_{4} \mathrm{Cl}\right]=\frac{\text { mole of } \mathrm{NH}_{4} C l}{\text { volume }}=\frac{2\; \mathrm{mmol}}{(40+20) \mathrm{ml}}=\frac{1}{20} \mathrm{~mol} / \mathrm{L}}

\mathrm{p H=7-\frac{1}{2} \times 5-\frac{1}{2} \log \left(\frac{1}{30}\right)}

\mathrm{pH=7-\frac{5}{2}+\frac{1}{2}[log\; 5+log\; 3+log\; 2]}

\mathrm{pH=7-\frac{5}{2}+\frac{1}{2}[0.65+0.48+0.30]}

\mathrm{pH=7-2.5+0.735}

\mathrm{pH=5.235}

Hence, the correct answer is Option (3).

Posted by

Rishi

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