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  • <img alt="50 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}" src="https://entrancecorner.oncodecogs.com/gif.latex?50%20%5Cmathrm%7B%7EmL%7D%24%20of%20%240.1%20%5Cmathrm%7BM%7D%20%5C

50 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH} is being titrated against 0.1 \mathrm{M} \mathrm{NaOH}$. When $25 \mathrm{~mL} of \mathrm{NaOH} has been added, the \mathrm{pH} of the solution will be__________ \times 10^{-2}. (Nearest integer)

(Given : \left.\mathrm{pK}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=4.76\right)

\begin{aligned} &\log 2=0.30 \\ &\log 3=0.48 \\ &\log 5=0.69 \\ &\log 7=0.84 \\ &\log 11=1.04 \end{aligned}

Option: 1

476


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{Given, \mathrm{pK}_{a}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=4.76}

\mathrm{moles\: of\; \mathrm{CH}_{3} \mathrm{COOH}=5 \mathrm{~m} \; mole}
\mathrm{moles \; of\; \mathrm{NaOH}=2.5 \mathrm{~m}\; mole}

Chemical Reaction -

                \mathrm{NaOH+CH_{3}COOH\longrightarrow CH_{3}COONa+H_{2}O}
\mathrm{t= 0 \quad 2.5 \, m mol \quad 5\, m mol \; \; \quad \quad \quad0 }
\mathrm{t= t \quad 0 \quad \quad \quad\quad 2.5\, m mol \; \; \quad \quad2.5\, m mol }

So buffer is formed whose pH is given as 

\mathrm{pH = pKa+log\left(\frac{[Salt]}{[Acid]}\right)}

\mathrm{\Rightarrow p{H}=p{Ka}+\log \left(\frac{2.5/75}{2.5 / 75}\right)}
\mathrm{\Rightarrow p{H}=p{K a}+\log (1)}
\mathrm{\Rightarrow p{H}=p{K a}}
\mathrm{\Rightarrow p{H}=4.76=476 \times 10^{-2}}

Hence 476 is the answer.

Posted by

manish

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