When 0.2 M solution of acetic acid is neutralized with 0.2 M NaOH in 500 mL of water, the pH of the resulting solution will be: [ p Ka. of acetic acid .=4.74]

Question

When 0.2 M solution of acetic acid is neutralized with 0.2 M NaOH in 500 mL of water, the pH of the resulting solution will be: [ p Ka. of acetic acid .=4.74] (A) 12.67 (B) 7.87 (C) 8.87 (D) 7

Tardigrade Admin · Accepted Answer

CH 3 COOH + NaOH arrow CH 3 COONa + H 2 O ⇒ At equivalence point, a salt of weak acid, strong base is formed ⇒ pH =7+(1/2)( p Ka+ log C) Here C= concentration of salt =(0.2 × 500/500+500)=0.1 M [Check that 500 mL of CH 3 COOH is also required] ⇒ pH =7+(1/2)(4.74+ log 0.1)=8.87

Q. When $0.2 M$ solution of acetic acid is neutralized with $0.2 M N a O H$ in $500 m L$ of water, the pH of the resulting solution will be: $[p K_{a}$ of acetic acid $= 4.74]$

$12.67$

$7.87$

$8.87$

7

Q. When 0.2M solution of acetic acid is neutralized with 0.2MNaOH in 500mL of water, the pH of the resulting solution will be: [pKa​ of acetic acid =4.74]

12.67

7.87

8.87

7

CH3​COOH+NaOH→CH3​COONa+H2​O ⇒ At equivalence point, a salt of weak acid, strong base is formed ⇒pH=7+21​(pKa​+logC) Here C= concentration of salt =500+5000.2×500​=0.1M [Check that 500mL of CH3​COOH is also required] ⇒pH=7+21​(4.74+log0.1)=8.87

Q. When $0.2 M$ solution of acetic acid is neutralized with $0.2 M N a O H$ in $500 m L$ of water, the pH of the resulting solution will be: $[p K_{a}$ of acetic acid $= 4.74]$

$12.67$

$7.87$

$8.87$