The pH of 0.1 M acetic acid solution is closest to [Dissociation constant of the acid, Ka=1.8 × 10-5 ]

Question

The pH of 0.1 M acetic acid solution is closest to [Dissociation constant of the acid, Ka=1.8 × 10-5 ] (A) 2.87 (B) 1.00 (C) 2.07 (D) 4.76

Tardigrade Admin · Accepted Answer

Given, dissociation constant, Ka=18 × 10-5 Concentration of [ H +]=0.1 M For dissociation of acetic acid CH 3 COOH leftharpoons CH 3 COO -+ H + H + =√ka c =√1.8 × 10-5 × 0.1 =√1.8 × 10-3 =1.34 × 10-3 pH =- log [ H +] =- log (1.34 × 10-3) =3- log 1.34 =2.87

Q. The pH of $0.1 M$ acetic acid solution is closest to [Dissociation constant of the acid, $K_{a} = 1.8 \times 1 0^{- 5}$ ]

$2.87$

$1.00$

$2.07$

$4.76$

Q. The pH of 0.1M acetic acid solution is closest to [Dissociation constant of the acid, Ka​=1.8×10−5 ]

2.87

1.00

2.07

4.76

Given, dissociation constant, Ka​=18×10−5 Concentration of [H+]=0.1M For dissociation of acetic acid CH3​COOH⇌CH3​COO−+H+ H+=ka​c​ =1.8×10−5×0.1​ =1.8​×10−3 =1.34×10−3 pH=−log[H+] =−log(1.34×10−3) =3−log1.34 =2.87

Q. The pH of $0.1 M$ acetic acid solution is closest to [Dissociation constant of the acid, $K_{a} = 1.8 \times 1 0^{- 5}$ ]

$2.87$

$1.00$

$2.07$

$4.76$