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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%.

PBCHEM

Answer : 1.91 %

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

1.  HCOOH ⇄ [tex]HCOO^{-} + H^{+} [/tex]      

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

As the equilibrium concentration of [tex] H^{+} [/tex] will be the pH of the solution.

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

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

3.  The moles present will be as

         [HCOOH] (M)                 [tex] [H^{+}] [/tex](M)           [tex][HCOO^{-}] [/tex](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 [tex] H^{+}[/tex] comes from the acid, and none from water.  

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

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

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 [tex] 10^{-3})^{2}[/tex] /0.50= 1.82 x [tex] 10^{-4}[/tex]


Now, calculating the percent ionization for this problem.  

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


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

∴ percent ionization = {(9.55 x [tex] 10^{-3})[/tex]/ (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%.

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