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?
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.
To find the value of Ka for the weak acid HA, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Where pH is the given value of 2.200, pKa is the negative logarithm of Ka, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
To find the concentrations, we first need to calculate the moles of HA and the volume of the solution.
Moles of HA = mass / molar mass
= 2.20 g / 65.0 g/mol
= 0.0338 moles
Now, we need to find the concentration of HA in moles per liter (M):
Concentration of HA = moles / volume
= 0.0338 moles / 0.750 L
= 0.045 moles/L
Since HA is a weak acid, it will dissociate partially, so the concentration of A- is equal to the concentration of HA. Therefore, [A-] = [HA] = 0.045 moles/L.
Now, we can substitute the values into the Henderson-Hasselbalch equation:
2.200 = pKa + log (0.045/0.045)
Since the concentration of [A-] is equal to the concentration of [HA], their ratio is 1, so the log term simplifies to 1.
2.200 = pKa + 1
Rearranging the equation:
pKa = 2.200 - 1
pKa = 1.200
Finally, to find the value of Ka, we need to take the antilog of the pKa:
Ka = 10^(pKa)
Ka = 10^(1.200)
Ka ≈ 15.85
Therefore, the value of Ka for the weak acid HA is approximately 15.85.