a 8.6 *10 to the -3 power M solution of H3PO4 has a PH of 2.30 what is the Ka for H3PO4?
Convert pH to (H^+). That's approximately 5E-3.
............H3PO4 ==> H^+ + H2PO4^-
I.........0.0086 M....0.......0
C...........-x........x........x
E.....0.0086-x........x........x
The problem tells you x = (H^+) = (approximately) 5E-3
Plug that in for x (into the Ka expression) and solve for Ka.
To find the Ka for H3PO4, we can use the equation for the dissociation of a weak acid and the given pH value.
The dissociation equation for H3PO4 is:
H3PO4 ⇌ H+ + H2PO4-
From this equation, we can see that for every 1 mol of H3PO4, 1 mol of H+ ions and 1 mol of H2PO4- ions are produced. Therefore, the initial concentration of H3PO4 will be the same as the concentration of H+ ions.
Given that the concentration of the H3PO4 solution is 8.6 x 10^(-3) M, the initial concentration of H+ ions is also 8.6 x 10^(-3) M.
Now, to calculate the Ka for H3PO4, we need to use the formula for pH:
pH = -log[H+]
Since we know the pH is 2.30, we can rearrange the formula to solve for [H+]:
[H+] = 10^(-pH)
Substituting the given pH value:
[H+] = 10^(-2.30)
Calculating, we find:
[H+] = 0.00498 M (approximately)
Since the initial concentration of H3PO4 is equal to the concentration of H+ ions:
[H3PO4] = 0.00498 M
Now, we can use the equation for Ka:
Ka = ([H+][H2PO4-]) / [H3PO4]
Substituting the known values:
Ka = (0.00498 * 0.00498) / 0.00498
Simplifying:
Ka = 0.00498
Therefore, the Ka for H3PO4 is 0.00498.