Question

A 0.50 M solution of formic acid, HCOOH, has a pH of 2.02. Calculate the percent ionization of HCOOH

Answer 1

Answer is: the percent ionizationof formic acid is 1,82%.
Chemical reaction: HCOOH(aq) ⇄ H⁺(aq) + HCOO⁻(aq).
pKa(HCOOH) = 3,77.

Ka(HCOOH) = 1,7·10⁻⁴.

c(HCOOH) = 0,5 M.

[H⁺] = [HCOO⁻] = x; equilibrium concentration.
[HA] = 0,1 M - x.
Ka = [H⁺] · [HCOO⁻] / [HCOOH].
0,00017 = x² / 0,5 M - x.
Solve quadratic equation: x = 0,0091 M.
α = 0,0091 M ÷ 0,5 M · 100% = 1,82%.

Answer 2

Answer : 1.91 %

Explanation : The steps to solve this problem are explained below;

1.  HCOOH ⇄ $HCOO^{-} + H^{+}$      

Here Ka =($[HCOO^{-}]_{eq} X [H^{+}]_{eq}$ )/  $[HCOOH]_{eq}$                                                        

As the equilibrium concentration of $H^{+}$ will be the pH of the solution.

∴ $[H^{+}]_{eq}$ = $10^{(-2.02)} = 9.55 x [tex] 10^{-3}$ M  

2. The initial concentration of HCOOH.  When it loses x moles from it as the acid undergoes dissociation to form $HCOO^{-}$ and $H^{+}$.  

3.  The moles present will be as

         [HCOOH] (M)                 $[H^{+}]$(M)           $[HCOO^{-}]$(M)  

Initial       0.50                                0.00                                          0.00

After Change  -x                              +x                                            +x

Equilibrium      ( 0.50 -x)                     x                                              x

∴ Ka   =   (x) x (x)  / (0.50 - x)  

4.  Assuming that all of the $H^{+}$ comes from the acid, and none from water.  

As $[H^{+}]_{eq}$ = 9.55 x$10^{-3}$ which is much higher than the 1.0 x$10^{-7 }$ M [tex[H^{+}[/tex] from water.  

Also, the concentration of HCOOH will change very little, from 0.50  to 0.50 - 9.55 x $10^{-3}$.  

The change in concentration can be ignored if it is less than 5% of the original concentration.  

∴ 0.50 M x 5% = 0.025, so the change in [HCOOH] in this problem can be ignored.    

Now,  Ka = (x)(x)/0.50 = (9.55 x $10^{-3})^{2}$ /0.50= 1.82 x $10^{-4}$

Now, calculating the percent ionization for this problem.  

which will represent the relative number of acid molecules which dissociate. It is calculated as :

$[H^{+}]_{eq}$ x  100 /$[HCOOH]_{i}$                                                        

∴ percent ionization = {(9.55 x $10^{-3})$/ (0.50)}x 100 = 1.91 %

This value of 1.91 % indicates that very little of this acid dissociates (ionizes) under these conditions.  

For strong acids and bases, the percent ionization is 100%.