Question

What is pH of 50 ml 0.05 (M) NaOH and 50 ml 0.1 (N) $C{H_3}COOH$.

Answer 1

Hint: First one should know that pH of the total solution will be determined using the product of above reactants which is a sodium salt of carboxylic acid. Further, we know that
Number of moles = Molarity of the solution $ \times $Volume of the solution (in L)
Using this, we will find the number of moles and then pH.
We know that ${K_a}(C{H_3}COOH) = 1.8 \times {10^{ - 5}}$

Complete step by step answer:
First, let us understand the reaction that is occurring between NaOH and $C{H_3}COOH$.
$C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O$
From the balanced equation, it is clear that one mole of NaOH reacts with one mole of $C{H_3}COOH$ to give salt and water. So, they react in 1:1 ratio.
Now, let us write the things given to us.
So, we have Molarity of NaOH = 0.05 M
Volume of NaOH = 50 mL = 0.05 L
Normality of $C{H_3}COOH$= 0.1 N
Volume of $C{H_3}COOH$= 50 mL = 0.05 L
Total Volume of mixture after combining the $C{H_3}COOH$and NaOH is-
Total volume of mixture = Volume of $C{H_3}COOH$+ Volume of NaOH
Total volume of mixture = 50 + 50 = 100 mL = 0.1 L
Now, let us calculate the number of moles.
Number of moles = Molarity of the solution $ \times $Volume of the solution (in L)
Number of moles of NaOH = 0.05 M $ \times $ 0.05 L
Number of moles of NaOH = 0.0025
Number of moles of $C{H_3}COOH$= 0.1$ \times $0.05 L
Number of moles of $C{H_3}COOH$= 0.005
Thus, 0.005 moles of $C{H_3}COOH$will react with NaOh giving 0.005 moles of $C{H_3}COONa$ in total 100 mL solution.
Thus, molarity of $C{H_3}COONa$ solution = $\dfrac{{Number{\text{ of moles}}}}{{Volume{\text{ of solution}}}}$
Thus, molarity of $C{H_3}COONa$ solution = $\dfrac{{0.005}}{{0.1}}$
Molarity of $C{H_3}COONa$ solution =0.05 M
We know that ${K_a}(C{H_3}COOH) = 1.8 \times {10^{ - 5}}$
We can find ${K_b} = \dfrac{{{{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 5}}}}$

${K_b} = 5.56 \times {10^{ - 10}}$
From the value of ${K_b}$, we can calculate $[O{H^ - }]$
${K_b} = \dfrac{{{{[O{H^ - }]}_2}}}{{[C{H_3}COONa]}}$
Putting the value of ${K_b}$ and $C{H_3}COONa$; we can find out value of $[O{H^ - }]$.
$5.56 \times {10^{ - 10}} = \dfrac{{{{[O{H^ - }]}_2}}}{{0.05}}$
${[O{H^ - }]_2} = 5.56 \times {10^{ - 10}} \times 0.05$
${[O{H^ - }]_2} = 2.78 \times {10^{ - 11}}$
$[O{H^ - }] = 5.27 \times {10^{ - 6}}$
Further, we know that pOH = - log $[O{H^ - }]$
So, pOH = -log$(5.27 \times {10^{ - 6}})$
pOH = 5.28
 We know, pH = 14-pOH
Thus, pH = 14-5.28
pH=8.72

Thus, the pH of the total solution is 8.72.

Note: During calculating the number of moles, the volume of solution is to be taken in L only. If we take the value in mL then the whole numerical will be wrong. Further, the pH of a system is negative logarithm of hydrogen ion concentration. The total value of pH and pOH comes out to be 14. So, the pH scale ranges from 0 to 14.