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 the solution? Calculate the additional volume x (in mL) of 0.2 M NaOH required to make pH of the solution 4.74. The ionisation constant of acetic acid is 1.8 × 10−5.
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 the solution? Calculate the additional volume x (in mL) of 0.2 M NaOH required to make pH of the solution 4.74. The ionisation constant of acetic acid is 1.8 × 10−5.
The correct option is B pH = 4.568, x = 4.83 mL
The addition of NaOH converts equivalent amount of acetic acid into sodium acetate.
So, Concentration of acetic acid after the addition of sodium hydroxide = 5070 × 0.2 M
Concentration of sodium acetate after the addition of sodium hydroxide = 2070 × 0.2 M
pH = pKa + logSaltacid- (1)
= −log(1.8 × 10−5) + log2050 = 4.745 - 0.397 = 4.348
Let x mL be the additional volume of 0.2 M NaOH that is need to be added to make pH of the solution equal to 4.74.
Substituting this value in Eq. (1), we get
log(20+x)(30−x) = −0.005
Solving for x , we have the required volume = 4.83 mL
pH = 4.568, x = 4.83 mL
The addition of NaOH converts equivalent amount of acetic acid into sodium acetate.
So, Concentration of acetic acid after the addition of sodium hydroxide = 5070 × 0.2 M
Concentration of sodium acetate after the addition of sodium hydroxide = 2070 × 0.2 M
pH = pKa + logSaltacid- (1)
= −log(1.8 × 10−5) + log2050 = 4.745 - 0.397 = 4.348
Let x mL be the additional volume of 0.2 M NaOH that is need to be added to make pH of the solution equal to 4.74.
Substituting this value in Eq. (1), we get
log(20+x)(30−x) = −0.005
Solving for x , we have the required volume = 4.83 mL
