I don't understand ice table at all..can anyone please guide me in how to solve the following question? :
The pH of a 0.00250 mol/L solution of benzoic acid is 3.65. Calculate the Ka for the benzoic acid. (This question does not state that the given concentration is an equilibrium concentration)
Thank u in advance
Benzoic acid is C6H5COOH. It ionizes as
C6H5COOH + H2O ==> H3O+ + C6H5COO^-
The ICE table runs from left to right in three horizontal rows:
I = initial...then the concns of each are listed.
C = change are listed in the row.
E = equilibrium listed in the row.
Unfortunately these boards don't allow spacing so I will use periods to space. Just ignore the periods.
pH = 3.65 so (H3O^+) = 2.24 x 10^-4 and I'm sure that is an equilibrium value.
C6N5COOH + H2O ==> H3O^+ + C6H5COO^-
I=0.00250...........0........0
C= -2.24E-4........+2.24E-4..+2.24E-4
E= 0.00250-2.24E-4...+2.24E-4...+2.24E-4
Ka = ((H3O^+)(C6H5COO^-)/(C6H6COOH)
Plug in the equilibrium values and solve for Ka.
Thank u sooo much!! =)
Just to confirm, the answer is 2.004748935 x 10^-5, right?
To solve this question, you need to understand the concept of the ice table and how it is used to calculate equilibrium concentrations and constants for expressions.
First, let's write the balanced equation for the ionization of benzoic acid (HA):
HA ⇌ H+ + A-
Now, let's create an ice table. The ice table is a tool used to keep track of changes in concentrations during a chemical reaction at equilibrium. It stands for Initial, Change, and Equilibrium.
In this case, we are given the initial concentration of benzoic acid (HA) as 0.00250 mol/L, and the initial concentration of H+ and A- ions is 0. After ionization, x mol/L of HA dissociates to form x mol/L of H+ and x mol/L of A-.
Using the ice table, we can set up the following information:
HA ⇌ H+ + A-
Initial (mol/L) 0.00250 0 0
Change (mol/L) -x +x +x
Equilibrium (mol/L) 0.00250-x x x
The change column represents the change in concentration at equilibrium, while the equilibrium column gives the final concentration after the reaction.
The equation for Ka is given as Ka = [H+][A-]/[HA]. Since the initial concentration of H+ and A- is 0, we can substitute the equilibrium concentrations into the Ka expression:
Ka = (x)(x)/(0.00250-x)
We are also given that the pH of the solution is 3.65, which implies that the concentration of H+ ions is equal to 10^(-pH). So, [H+] = 10^(-3.65).
Substituting this value into the equation, we get:
Ka = (10^(-3.65))(10^(-3.65))/(0.00250-10^(-3.65))
Now, using a calculator, evaluate this expression to find the value of Ka.
Note: In this problem, we assume that the dissociation of benzoic acid is small enough that the equilibrium concentration can be approximated as the initial concentration minus x. However, if the dissociation is significant, you may need to solve the quadratic equation resulting from the Ka expression.
I hope this step-by-step explanation helps you understand how to approach this question!