Asked by Sue
An unknown monoprotic weak acid, HA, has a molar mass of 65.0. A solution contains 2.20 g of HA dissolved 750. mL of solution. The solution has a pH of 2.200. What is the value of Ka for HA?
Answers
Answered by
DrBob222
HA ==> H^+ + A^-
(H^+)(A^-)/(HA) = Ka
moles HA = moles/molar mass = 2.20/65 = 0.0338
(HA) = (0.0338/0.750) = 0.0451 M
pH = 2.200
2.200 = -log(H^+)
(H^+) = 0.00631
Substitute into Ka expression.
(0.00631)(0.00631)/(0.0451-0.00631) = Ka.
You can finish.
(H^+)(A^-)/(HA) = Ka
moles HA = moles/molar mass = 2.20/65 = 0.0338
(HA) = (0.0338/0.750) = 0.0451 M
pH = 2.200
2.200 = -log(H^+)
(H^+) = 0.00631
Substitute into Ka expression.
(0.00631)(0.00631)/(0.0451-0.00631) = Ka.
You can finish.
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