A 8.6 x 10^-3M solution of H3PO4 has a pH=2.30. What is Ka for H3PO4
H3PO4 has three ionization constants, k1, k2, and k3. Probably you mean to calculate k1.
H3PO4 ==> H^+ + H2PO4^-2
k1 = (H^+)(H2PO4^-2)/(H3PO4)
Convert pH = 2.30 to (H^+). Assume (H2PO4^-2) is the same (almost true). Prepare an ICE chart, substitute and calculate k1. Post your work if you get stuck.
Ka = 1.8 X 10-4
To determine the Ka for H3PO4, we can use the given pH value and concentration of the acid.
Ka, also known as the acid dissociation constant, measures the extent to which an acid dissociates in water. It can be calculated using the equation:
Ka = [H+][A-]/[HA]
Where [H+] refers to the concentration of hydrogen ions, [A-] refers to the concentration of conjugate base ions, and [HA] refers to the concentration of the weak acid.
In this case, we are given the pH of the solution, which is 2.30. The pH is a measure of the concentration of hydrogen ions in a solution. It is calculated using the formula:
pH = -log[H+]
To find the concentration of hydrogen ions, we need to take the antilog of the negative pH value:
[H+] = 10^(-pH)
Substituting the given pH value:
[H+] = 10^(-2.30)
[H+] ≈ 0.00501187 M
Since H3PO4 is a weak acid, it dissociates into three H+ ions and one HPO4^2- ion. Therefore, the concentration of [A-] (HPO4^2-) would be three times the concentration of [H+] (after dissociation).
[A-] = 3 × [H+]
[A-] ≈ 3 × 0.00501187 M
[A-] ≈ 0.0150356 M
The concentration of [HA] (H3PO4) is given as 8.6 × 10^-3 M.
Now, we can substitute these values into the Ka equation:
Ka = [H+][A-]/[HA]
Ka = (0.00501187 M) × (0.0150356 M) / (8.6 × 10^-3 M)
Performing the calculations:
Ka ≈ 0.079
Therefore, the approximate value of Ka for H3PO4 is 0.079.