0.25 mol of formic acid (H C O2 H) is dissolved in enough water to make one litre of solution. The pH of that solution is 2.19. The Ka of formic acid is

Question

0.25 mol of formic acid (H C O2 H) is dissolved in enough water to make one litre of solution. The pH of that solution is 2.19. The Ka of formic acid is (A) 6.5× 10- 3 (B) 4.3× 10- 4 (C) 1.7× 10- 4 (D) 5.3× 10- 2

Tardigrade Admin · Accepted Answer

pH =2.19=-log [.H+]. [H+]=10- 2.19= Antilog (- 2.19) =6.46× 10- 3M Ka(H C O O H)=([H+] [H C O O-]/[H C O O H]) Here [H+]=[H C O O-]=6.46× 10- 3M [H C O O H]=0.25 mol/1L =0.25 text M and Ka(H C O O H)=? So mathrmK mathrma( mathrmHCOOH)=((6.46 × 10-3)2/0.25)=1.7 × 10-4.

Q. 0.25 mol of formic acid $(H C O_{2} H)$ is dissolved in enough water to make one litre of solution. The pH of that solution is 2.19. The $K_{a}$ of formic acid is

$6.5 \times 1 0^{- 3}$

$4.3 \times 1 0^{- 4}$

$1.7 \times 1 0^{- 4}$

$5.3 \times 1 0^{- 2}$

Q. 0.25 mol of formic acid (HCO2​H) is dissolved in enough water to make one litre of solution. The pH of that solution is 2.19. The Ka​ of formic acid is

6.5×10−3

4.3×10−4

1.7×10−4

5.3×10−2

pH =2.19=−log[H+] [H+]=10−2.19= Antilog (−2.19) =6.46×10−3M Ka​(HCOOH)=[HCOOH][H+][HCOO−]​ Here [H+]=[HCOO−]=6.46×10−3M [HCOOH]=0.25 mol/1L =0.25 M and Ka​(HCOOH)=? So Ka​(HCOOH)=0.25(6.46×10−3)2​=1.7×10−4.

Q. 0.25 mol of formic acid $(H C O_{2} H)$ is dissolved in enough water to make one litre of solution. The pH of that solution is 2.19. The $K_{a}$ of formic acid is

$6.5 \times 1 0^{- 3}$

$4.3 \times 1 0^{- 4}$

$1.7 \times 1 0^{- 4}$

$5.3 \times 1 0^{- 2}$