pH of a buffer prepared by adding 10 mL of 0.10 M acetic acid to 20 mL of 0.1 M sodium acetate is equal to ( p Ka. of .CH 3 COOH =4.74)

Question

pH of a buffer prepared by adding 10 mL of 0.10 M acetic acid to 20 mL of 0.1 M sodium acetate is equal to ( p Ka. of .CH 3 COOH =4.74) (A) 4.74 (B) 5.6 (C) 5.04 (D) 6.02

Tardigrade Admin · Accepted Answer

We can easily calculate the pH by converting the concentration into millimoles. Millimoles of CH 3 COOH =0.1 × 10=1.0 millimole Millimoles of CH 3 COONa =0.1 × 2.0=2.0 millimole ∴ Using Henderson-Hassebalch equation, pH = p Ka+ log ( text [Conjugate base ]/[ text acid ]) =4.74+ log (2/1) =4.74+0.30 =5.04

Q. $p H$ of a buffer prepared by adding $10 m L$ of $0.10 M$ acetic acid to $20 m L$ of $0.1 M$ sodium acetate is equal to $(p K_{a}$ of $C H_{3} COO H = 4.74)$

$4.74$

$5.6$

$5.04$

$6.02$

Q. pH of a buffer prepared by adding 10mL of 0.10M acetic acid to 20mL of 0.1M sodium acetate is equal to (pKa​ of CH3​COOH=4.74)

4.74

5.6

5.04

6.02

We can easily calculate the pH by converting the concentration into millimoles. Millimoles of CH3​COOH=0.1×10=1.0 millimole Millimoles of CH3​COONa=0.1×2.0=2.0 millimole ∴ Using Henderson-Hassebalch equation, pH=pKa​+log[ acid ] [Conjugate base ]​ =4.74+log12​ =4.74+0.30 =5.04

Q. $p H$ of a buffer prepared by adding $10 m L$ of $0.10 M$ acetic acid to $20 m L$ of $0.1 M$ sodium acetate is equal to $(p K_{a}$ of $C H_{3} COO H = 4.74)$

$4.74$

$5.6$

$5.04$

$6.02$