Question

$0.1mole$ each of $C{H}_{3}COOH$ and $C{H}_{3}COONa$ are dissolved in pure water to prepare $100mk$ of its solution. $p{K}_a$ of $C{H}_{3}COOH$ is $4.8$. How many milli moles of $NaOH$ to be added to the above solution to make its $pH=5.1$?

• A. $0.33$
• B. $33$
• C. $0.066$
• D. $66$

Answer 1

The correct option is B $33$

Given,

Initial concentration of $C{H}_{3}COOH=0.1$ mole in $100mL$ solution

$=\frac{0.1mole}{0.1L}=1mole{L}^{-1}$ $[Molarity=\frac{mol}{L}]$

Initial concentration of $C{H}_{3}COONa=0.1$ mole in $100mL$ solution

$=\frac{0.1mole}{0.1L}=1mole{L}^{-1}$

The reaction involved is

$C{H}_{3}COOH+NaOH⟶C{H}_{3}COONa+H_{2}O$

$C{H}_{3}COOH⟶C{H}_{3}CO{O}^{-}+H^{+}$

When $x$ mole per $100mL$ solution is added to initial solution of $NaOH$

New concentration of $C{H}_{3}COOH=1-(x\times 10)$ $⟶1$

New concentration of $C{H}_{3}COOna=1+(x\times 10)$ $⟶2$ $[\because \frac{xmole}{0.1L}]$

Now, as this is an acid buffer solution

$pH=p{K}_a+\log \frac{\left[Salt\right]}{\left[Acid\right]}$ can be used $⟶3$

We are given $pH$ of solution=$5.1$

$p{K}_a$ of $C{H}_{3}COOH=4.8$

From $1,2,3$

$5.1=4.8+\log \frac{[1+10x]}{[1-10x]}$

$5.1=4.8+\log \frac{[1+10x]}{[1-10x]}$

$\Rightarrow =0.3=log\frac{[1+10x]}{[1-10x]}$

Antilog $\left[0.3\right]=\frac{[1+10x]}{[1-10x]}$

$2=\frac{[1+10x]}{[1-10x]}$ $[\because \log 2=0.3010]$

$2120x=1+10x$

$\Rightarrow 30x=1$

$\Rightarrow x=\frac{1}{30}$

$=0.033$

$\therefore$ The no. of moles of $NaOH$ added=$x=0.033$ moles

$0.033$ moles=$33$ millimoles

$=33\times {10}^{-3}$ moles= $33$ millimoles