The pH of 0.01 M HCOOH will be. [Given: √ K a =1.18 × 10-2, . log 10 1.18=0.0719]

Question

The pH of 0.01 M HCOOH will be. [Given: √ K a =1.18 × 10-2, . log 10 1.18=0.0719] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

For weak acid, [ H +]=√ K a × C =√ K a × √ C =1.18 × 10-2 × √0.01 =1.18 × 10-2 × √10-2 =1.18 × 10-2 × 10-1 =1.18 × 10-3 pH = log 10[ H +] =- log 10[1.18 × 10-3] =-[ log 10 1.18+ log 10(10-3)] =-[0.0719-3]=-[-2.9281]=2.93

Q. The pH of 0.01MHCOOH will be_____. [Given: Ka​​=1.18×10−2,log10​1.18=0.0719]

For weak acid, [H+]=Ka​×C​ =Ka​​×C​ =1.18×10−2×0.01​ =1.18×10−2×10−2​ =1.18×10−2×10−1 =1.18×10−3 pH=log10​[H+] =−log10​[1.18×10−3] =−[log10​1.18+log10​(10−3)] =−[0.0719−3]=−[−2.9281]=2.93

Q. The $p H$ of $0.01 M H COO H$ will be_____.
[Given: $K_{a} = 1.18 \times 1 0^{- 2}, lo g_{10} 1.18 = 0.0719]$