pH of a buffer solution decreases by 0.02 units when 0.12 g of acetic acid is added to 250 mL of a buffer solution of acetic acid and potassium acetate at 27° C. The buffer capacity of the solution is

Question

pH of a buffer solution decreases by 0.02 units when 0.12 g of acetic acid is added to 250 mL of a buffer solution of acetic acid and potassium acetate at 27° C. The buffer capacity of the solution is (A) 0.1 (B) 10 (C) 1 (D) 0.4

Tardigrade Admin · Accepted Answer

Buffer capacity, β=(d C HA /d pH ), where, d C HA = no. of moles of acid added per litre ∴ d pH = change in pH d C HA =( text moles of acetic acid / text volume ) =(0.12 / 60/250 / 1000)=(1/125) ∴ β =(1 / 125/0.02)=(1/2.5)=0.4

Q. $p H$ of a buffer solution decreases by $0.02$ units when $0.12 g$ of acetic acid is added to $250 m L$ of a buffer solution of acetic acid and potassium acetate at $2 7^{\circ} C$ . The buffer capacity of the solution is

0.1

10

1

0.4

Buffer capacity, $β = \frac{d C _{H A}}{d _{p H}}$ ,

where, $d C_{H A} =$ no. of moles of acid added per litre

$∴ d_{p H} =$ change in $p H$

$d C_{H A} = \frac{moles of acetic acid}{volume}$

$= \frac{0.12/60}{250/1000} = \frac{1}{125}$

$∴ β = \frac{1/125}{0.02} = \frac{1}{2.5} = 0.4$

Q. pH of a buffer solution decreases by 0.02 units when 0.12g of acetic acid is added to 250mL of a buffer solution of acetic acid and potassium acetate at 27∘C. The buffer capacity of the solution is

0.1

10

1

0.4

Buffer capacity, β=dpH​dCHA​​, where, dCHA​= no. of moles of acid added per litre ∴dpH​= change in pH dCHA​= volume moles of acetic acid ​ =250/10000.12/60​=1251​ ∴β=0.021/125​=2.51​=0.4

Q. $p H$ of a buffer solution decreases by $0.02$ units when $0.12 g$ of acetic acid is added to $250 m L$ of a buffer solution of acetic acid and potassium acetate at $2 7^{\circ} C$ . The buffer capacity of the solution is

Buffer capacity, $β = \frac{d C _{H A}}{d _{p H}}$ ,

where, $d C_{H A} =$ no. of moles of acid added per litre

$∴ d_{p H} =$ change in $p H$

$d C_{H A} = \frac{moles of acetic acid}{volume}$

$= \frac{0.12/60}{250/1000} = \frac{1}{125}$

$∴ β = \frac{1/125}{0.02} = \frac{1}{2.5} = 0.4$