Question

20 ml of 0.2 M NaOH is added to 50 ml of 0.2 M acetic acid to give 70 ml of the solution. What is the pH of this solution? Calculate the additional volume of 0.2 M NaOH required to make the pH of the solution 4.74. the ionization constant of acetic acid is $1.8 \times {10^{ - 5}}$ :
A. pH=3.56, volume=5.86 ml
B. pH=3.56, volume=4.86 ml
C. pH=4.56, volume=4.86 ml
D. pH=4.56, volume=3.86 ml

Answer 1

Hint: We can clearly observe that the salt made above is due to the neutralization reaction between the strong base that is sodium hydroxide and the weak base( acetic acid). When a salt is dissolved in a solvent, it first dissociates into its constituent ions. This process is called dissolution.

Complete step by step solution:
The concept used in the question will be determination of pH due to hydrolysis of salt, it is important that one should be aware of the fact that whenever the salt of a weak acid and strong base is dissolved in water, the solution is basic.
Hence an acidic buffer gets formed and from henderson’s equation we know that
$pH = p{K_a} + \log \dfrac{{millimoles\, of\, salt}}{{millimoles \,of\, acid}}$ (Where ionization constant is $p{K_a}$ )
Where we know that millimoles is given by molarity*volume( in ml)
Therefore we are given with the molarity of both base and acid and hence we can calculate the millimoles correspondingly. On substituting the values we get,
Millimoles of NaOH = 0.2*20=4
Millimoles of acetic acid = 0.2*50=10
Millimoles of the salt produced = 4 and the remaining millimoles of acetic acid is (10-4=6 millimoles) since 4 millimoles of acetic acid has reacted with 4 millimoles of base to produce 4 millimoles of salt. Now the Ph will be,
$pH = - \log (1.8 \times {10^{ - 5}}) + \log \dfrac{4}{6}$ = 4.56
Now since we want a pH of 4.74 then let the additional volume of NaOH be v ml,
Millimoles of NaOH will be = 0.2 v
Millimoles of acetic acid be = 6-0.2v
Millimoles of salt = 4+0.2v
Hence substituting the above values in henderson’s equation
$
\Rightarrow 4.74 = - \log (1.86 \times {10^{ - 5}}) + \log \dfrac{{4 + 0.2v}}{{6 - 0.2v}} \\
\Rightarrow 4.74 = 4.7447 + \log \dfrac{{4 + 0.2v}}{{6 - 0.2v}} \\
\Rightarrow - 0.0047 = \log \dfrac{{4 + 0.2v}}{{6 - 0.2v}} \\
Taking \;\; anti\log \dfrac{{4 + 0.2v}}{{6 - 0.2v}} \\
v = 4.86\,ml
$

Hence the correct option is C.

Note:
Now if these ions chemically react with water the process is called hydrolysis the salt hydrolysis may be defined as the reaction of cation or anion of the salt with water to produce acidic or basic solutions.