Question

When $50mL$ of $0.1M$ $NaOH$ is mixed with $50mL$ of $0.05M$ $C{H_3}COOH$ solution, pH becomes:
A. $1.602$
B. $12.39$
C. $4.74$
D. $8.72$

Answer 1

We can solve this problem by calculating the concentration of the solution after mixing of sodium hydroxide and acetic acid. We know that the number of moles can be calculated by multiplying molarity and volume , hence the number of moles $= Molarity \times Volume\left( L \right)$ and the formula for calculation of pH value is $pH = 14 - pOH$.

Complete step by step answer:
It is given that $50mL$ of $0.1M$ $NaOH$ is mixed with $50mL$ of $0.05M$ $C{H_3}COOH$ so $50mL$ of $NaOH$ will neutralises $50mL$ of $C{H_3}COOH$.On mixing sodium hydroxide with acetic acid we get the following reaction;
$C{H_3}COOH + NaOH \rightleftharpoons {H_2}O + C{H_3}COONa$
So let's first calculate number of moles of acetic acid and sodium hydroxide with the help of formula number of moles $= Molarity \times Volume\left( L \right)$;
Number of moles of $NaOH$ $= 0.1M \times 50 \times {10^{ - 3}}L$
Number of moles of $NaOH$ $= 0.1M \times 0.05 = 0.005$ moles
Number of moles of $C{H_3}COOH$ $= 0.05M \times 50 \times {10^{ - 3}}L$
Number of moles of $C{H_3}COOH$ $= 0.05M \times 0.05L = 0.0025$ moles
Total volume of mixture $= 50mL + 50mL = 100mL = 0.1L$
Hence the number of moles of excess $NaOH$ $= 0.005 - 0.0025 = 0.0025$ moles
Now the concentration of excess $NaOH$ $= \dfrac{{0.0025}}{{0.1}} = 0.025M$
The pOH value of the solution will be ${p^{OH}} = - \log [O{H^ - }]$
Now the pH of the excess sodium hydroxide would surpass the pH of $C{H_3}COONa$ obtained after hydrolysis. So $pH = 14 - pOH$ $= 14 - ( - \log 0.025) = 12.39$. So the option B is the correct answer to this problem that is when $50mL$ of $0.1M$ $NaOH$ is mixed with $50mL$ of $0.05M$ $C{H_3}COOH$ solution, pH becomes $12.39$. Hence the nature of the solution will be alkaline as it has high pH value.
So, the correct answer is “Option B”.

Note: We have approached this problem with the help of the concept that sodium hydroxide is a strong base which will neutralize the $C{H_3}COOH$ (a weak acid) completely. Hence to calculate the pH we need to find out the concentration of the solution. After calculating concentration of resultant solution we get the pH value with the help of formula $pH = 14 - pOH$.