Question

Acid Dissociation lab. Solution 1: 0.02M, pH3.20, solution 2: 0.2M pH3.67

Construct an ICE table and use the initial molarity of [HC2H3O2], the equilibrium concentration for [H3O+]eq and the equation;

HC2H3O2-(aq)+H2O(l)=H3O+(aq)+C2H3O2-(aq)

to determine HC2H3O2-]eq and [HC2H3O2]eq

Can you please break it down step by step to I can use these examples to do other ICE problems and get the system down, thank you!

Answer 1

Solution 1 :

As initial Conc. of HC2H3O2 = 0.02 M

Reaction :

HC2H3O2 + H2O <----> C2H3O2- + H3O+

initial 0.02 0 0

change - x + x +x

Equilibrium 0.02 - x x x

Now , pH = 3.20

So , [H3O+]eq = x = 10^-pH = 10^-3.20 = 0.00063 M

Therefore , Equilibrium concentration of all species are :

[HC2H3O2]eq = 0.02 - x = 0.02 - 0.00063 =  0.01937 M  

[C2H3O2-]eq = x =  0.00063 M  

Solution 2 :

As initial Conc. of HC2H3O2 = 0.2 M

Reaction :   

HC2H3O2 + H2O <----> C2H3O2- + H3O+

initial 0.2 0 0

change - x + x +x

Equilibrium 0.2 - x x x  

Now , pH = 3.67

So , [H3O+]eq = x = 10^-pH = 10^-3.67 = 0.00021   M

Therefore , Equilibrium concentration of all species are :

[HC2H3O2]eq = 0.2 - x = 0.2 - 0.00021 =  0.19979 M  

[C2H3O2-]eq = x = 0.00021 M