The pH of 0.05 M acetic acid is (Ka=2 × 10-5)

Question

The pH of 0.05 M acetic acid is (Ka=2 × 10-5) (A) 2 (B) 11 (C) 10-3 (D) 3

Tardigrade Admin · Accepted Answer

CH 3 COOH leftharpoons H ++ CH 3 COO - Ka=([ H +][ CH 3 COO -]/[ CH 3 COOH ])=([ H +]2/[ CH 3 COOH ]) [ H +] =√Ka[ CH 3 COOH ] =√2 × 10-5 × 0.05 =√10-6 [ H +] =10-3 ∵ pH =- log [ H +] =- log [10-3] ∴ pH =3

Q. The $p H$ of $0.05 M$ acetic acid is $(K_{a} = 2 \times 1 0^{- 5})$

2

11

$1 0^{- 3}$

3

Q. The pH of 0.05M acetic acid is (Ka​=2×10−5)

2

11

10−3

3

CH3​COOH⇌H++CH3​COO− Ka​=[CH3​COOH][H+][CH3​COO−]​=[CH3​COOH][H+]2​ [H+]=Ka​[CH3​COOH]​ =2×10−5×0.05​ =10−6​ [H+]=10−3 ∵pH=−log[H+] =−log[10−3] ∴pH=3

Q. The $p H$ of $0.05 M$ acetic acid is $(K_{a} = 2 \times 1 0^{- 5})$

$1 0^{- 3}$