A0.50 m solution of formic acid, hcooh, has a ph of 2.02. calculate the percent ionization of hcooh

Answer : 1.91 %

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

1.  HCOOH ⇄      

Here Ka =( )/                                                          

As the equilibrium concentration of will be the pH of the solution.

∴ = M  

2. The initial concentration of HCOOH.  When it loses x moles from it as the acid undergoes dissociation to form and .  

3.  The moles present will be as

         [HCOOH] (M)                 (M)           (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 comes from the acid, and none from water.  

As = 9.55 x which is much higher than the 1.0 x M [tex[H^{+}" alt=" 10^{-3}" /> M  

2. The initial concentration of HCOOH.  When it loses x moles from it as the acid undergoes dissociation to form and .  

3.  The moles present will be as

         [HCOOH] (M)                 (M)           (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 comes from the acid, and none from water.  

As = 9.55 x which is much higher than the 1.0 x M [tex[H^{+}" /> from water.  

Also, the concentration of HCOOH will change very little, from 0.50  to 0.50 - 9.55 x .  

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 /0.50= 1.82 x

Now, calculating the percent ionization for this problem.  

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

x  100 /                                                        

∴ percent ionization = {(9.55 x / (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%.

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.