What is the pH of a solution obtained by adding 15 g deK2SO4em 150 ml NaHSO4 with Molarity of 0, 2 m. HSO4 Ka = 1, 3.10^-2
[H+] fm Potassium Sulfate is an hydrolysis calculation using (15g/150ml) = 0.575M sulfate ion in 150 ml soln. Calculate equivalence point [H+](K2SO4) = 1.8 x 10^-8M
[H+] fm 0.20M Bisulfate ion is a weak acid in solution calculation in which [H+] is determined using the quadratic equation b/c Ka of the Bisulfate ion is quite large. [HSO-]/Ka < 100 and the 'x' in the I.C.E. table can not be dropped.
[H+](HSO4-) = 5.0 X 10-2M
[H+](K2SO4) = 1.5 X 10^-8
[H+](HSO4-) = 5.0 X 10^-2
___________________________
Sum [H+] = 0.05M (The presence of K2SO4 makes very little contribution to the [H+] and the dominant source for [H+] is from the Bisulfate ionization giving 5.0 x 10^2M in H+) >>> 1.5 x 10^-8.
Therefore, using [H+](sum) = 0.05M => pH = -log[H+] = -log(0.05) = -(-1.3) = 1.3
I think the correct solution is to recognize this as a buffer problem use the Hendersib-Hasselbalch equation.
pH = pK2 + log (base)/(acid)
pH = 1.88 + log (0.574/0.2) = ?
pH approx 2.3
OR you can work it as a common ion; i.e.,
HSO4^- ==> H^+ + SO4^- k2 = ?
K2SO4 ==> 2K^+ + SO4^2-
K2 = (H^+)(SO4^2)/(HSO4^-)
You know SO4^- is o.574
You know HSO4^- is 0.2M
Solve for H^+ = ?
This gives the same answer as the HH solution; it also allows you to use the quadratic equation. I did both and came out with 2.3 for the HH equation, 2.3 for the other and 2.4 if the second method uses the quadratic.
To find the pH of a solution obtained by adding deK2SO4 and NaHSO4, we need to consider the dissociation of the HSO4- ion and the resulting equilibrium.
First, let's understand the reaction:
NaHSO4 → Na+ + HSO4-
Since the concentration of NaHSO4 given is in molarity (0.2 M), we can directly consider the concentration of HSO4- as 0.2 M.
The Ka value given (1.3 x 10^-2) is the acid dissociation constant for HSO4-. It tells us the degree to which HSO4- dissociates in water.
The equation for the dissociation of HSO4- is as follows:
HSO4- ⇌ H+ + SO4^2-
To find the concentration of H+, we need to set up an ICE table and apply the Ka expression.
Let's assume the change in concentration of H+ is "x". Then, the change in concentration of SO4^2- will also be "x", and the initial concentration of HSO4- is 0.2 M. The initial concentration of H+ is 0.
Using the Ka expression, we have:
Ka = [H+][SO4^2-] / [HSO4-]
Substituting the known values, we get:
1.3 x 10^-2 = x*x / (0.2 - x)
Simplifying the equation further, we have:
1.3 x 10^-2 * (0.2 - x) = x^2
Now, let's solve this quadratic equation to find the value of "x". By solving the equation, we find that x ≈ 7.82 x 10^-3 M.
Since pH is defined as the negative logarithm of the concentration of H+, we can determine the pH by taking the negative base-10 logarithm of the concentration of H+:
pH = -log(H+)
pH ≈ -log(7.82 x 10^-3)
Using a calculator, we find that the pH is approximately 2.107.
Therefore, the pH of the solution obtained by adding 15 g deK2SO4 and 150 ml NaHSO4 with a molarity of 0.2 M is approximately 2.107.