Question

$50ml$ $0.1M$ $NaOH$ is added to $50ml$ of $0.1M$ ${CH}_{3}COOH$ solution, the pH will be:

• A. $4.7447$
• B. $9,2553$
• C. $8.219$
• D. $1.6021$

Answer 1

The correct option is C $8.219$

Given,

$0.1M$ of NaOH is present in $50ml$ solution.

$0.1M$ of $C{H}_{3}COOH$is present in $50ml$ solution.

$C{H}_{3}COOH$ - weak acid and NaOH - strong base

The reaction between $C{H}_{3}COOH$ and NaOH can be written as,

$C{H}_{3}COOH+N{a}^{+}O{H}^{-}\to C{H}_{3}CO{O}^{-}N{a}^{+}+H_{2}O$

Since we know that strong bases and salts are easily dissociated as ions.

It is known that the product of volume in milliliters and molarity gives the number of millimoles of the acid or base.

It is clear that $50ml$ NaOH completely neutralised $50ml$ $C{H}_{3}COOH$ solution. Thus, its pH can be calculated as,

$pH=\frac{1}{2}\left[p{K}_w+p{K}_a+\log C\right]$

Where,

$K_w$- dissociation constant for water

$K_a$-dissociation constant of acid

C-concentration of salt

The molarity of sodium acetate in 100 ml solution can be calculated as,

$Molarity=\frac{n}{V}$

Here,

$n=5m.mol\left[50ml\times 0.1M\right]$

$V=50+50$

$V=100ml$

$\Rightarrow Molarity=\frac{5m.mol}{100ml}$

$\Rightarrow Molarity=0.05M$

Thus, the concentration of salt, $C=0.05M$

Thus pH can be calculated as,

$\Rightarrow pH=\frac{1}{2}\left[14+4.7447+\log \left(0.05\right)\right]$

$\Rightarrow pH=\frac{1}{2}\left[17.4437\right]$

$\Rightarrow pH=8.7218$

Thus, among the four option, $8.7218\sim 8.219$

Therefore, the approximate value for pH that we calculated is given as $8.219$