A given weak acid HZ has a Ka=1.7×10−6.
What is the H3O+ concentration of a solution of HZ that has a concentration of 0.78 mol/L?
What is the pH of this solution of HZ?
What is the percent ionization of HZ in this solution?
Answer in units of %
........HZ + H2O ==> H3O^+ + Z^-
I......0.78M.........0........0
C........-x..........x........x
E.....0.78-x.........x........x
Ka = (H3O^+)(Z^-)/(HZ)
Use Ka, plug in E line of ionization above and solve for x = (H3O^+).
Then pH = -log(H^3O^+)
%ion = [(H3O^+)/0.78]*100 = ?
Post your work if you get stuck.
I got the first two parts correct but I can't seem to get the last part. The concentration I got for H3O^+ is 1.151520734e-3. When I plugged that in for the %ion I got 0.1476308633. I typed that into my homework and it said that is incorrect.
I have the same answer you obtained BUT my best guess is that you have reported too many significant figures. You are allowed only two; therefore, I would round that off to 0.15%. Let me know if that clears up the problem (or doesn't clear up the problem).
I tried .15 and it said it was incorrect.
To find the H3O+ concentration of a solution of HZ, first, we need to understand that the concentration of H3O+ is equal to the concentration of HZ that has ionized. Since HZ is a weak acid, it does not completely ionize in water.
Let's define the concentration of H3O+ as [H3O+], and the concentration of HZ as [HZ]. Since the initial concentration of HZ is 0.78 mol/L, the concentration of H3O+ in the solution will be equal to [HZ] at equilibrium.
Now, let's consider the equilibrium equation for the ionization of HZ:
HZ ⇌ H+ + Z-
The equation for the ionization constant expression, Ka, is:
Ka = [H+][Z-] / [HZ]
Since we know the value of Ka (1.7×10−6), we can write the equation as:
1.7×10−6 = [H+][Z-] / 0.78
Since the concentration of [H+]=[Z-] at equilibrium (as mentioned before), we can replace [H+] and [Z-] with [HZ]:
1.7×10−6 = [HZ]^2 / 0.78
Rearranging the equation:
[HZ]^2 = 1.7×10−6 × 0.78
Taking the square root of both sides:
[HZ] ≈ √ (1.7×10−6 × 0.78)
Calculating the value:
[HZ] ≈ 0.00107 mol/L
Therefore, the concentration of H3O+ in the solution is approximately 0.00107 mol/L.
To find the pH of the solution, we need to calculate the negative logarithm of the H3O+ concentration:
pH = -log[H3O+]
Using the H3O+ concentration found earlier:
pH = -log(0.00107)
Calculating the value:
pH ≈ 2.97
Therefore, the pH of the solution is approximately 2.97.
To find the percent ionization of HZ, we need to calculate the ratio of the ionized concentration to the initial concentration of HZ and multiply it by 100.
Percent Ionization = ([H3O+] / [HZ]) * 100
Using the values we obtained earlier:
Percent Ionization = (0.00107 / 0.78) * 100
Calculating the value:
Percent Ionization ≈ 0.137 * 100
Therefore, the percent ionization of HZ in this solution is approximately 13.7%.