Question

What volume of 0.1 M CH3COOH is to be added to 50 ml of 0.2 M Na-acetate to prepare a buffer solution of pH=4.91

Answer 1

your complete question is ---> What volume of 0.1 M CH3COOH is to be added to 50 ml of 0.2 M Na-acetate to prepare a buffer solution of pH=4.91 given pKa for acetic acid is 4.76.

solution : Let volume of acetic acid is V mL

given, concentration of acetic acid = 0.1M

so, number of mole of acetic acid = 0.1V milimole

concentration of sodium acetate = 0.2M

and volume of sodium acetate solution= 50mL

so, number of mole of sodium acetate = 0.2 × 50 = 10 milimole

after mixing,

concentration of acetic acid, [acid] = 0.1V/(V + 50) M

concentration of sodium acetate, [salt] = 10/(V + 50) M

now using formula

pH = pKa + log[salt]/[acid]

4.91 = 4.76 + log[10/0.1V]

or, 4.91 - 4.76 = log(100/V)

or, 0.15 = log(100/V)

or, 10^{0.15} = 100/V

or, V = 100/10^{0.15} ≈ 71mL

hence, 71 mL of acetic acid should be added to sodium acetate to prepare buffer solution of pH = 4.91

Answer 2

Answer:

1oo ml anno answer varunathh aaa enik ariyila atha