(a) First, we need to find the concentration of H^+ ions from the given pH value:
pH = -log10[H^+]
4.80 = -log10[H^+]
[H^+] = 10^(-4.80) = 1.58 x 10^(-5) M
Now, we can use the expression for Ka to find the pKa of the acid:
Ka = (H^+)(A^-)/(HA)
(HA) = 0.12 mol
(A^-) = 0.08 mol
Ka = (1.58 x 10^(-5))(0.08)/(0.12) = 1.05 x 10^(-5)
Now, we can find the pKa:
pKa = -log10(Ka) = -log10(1.05 x 10^(-5)) = 4.98
So, the pKa of the acid is 4.98.
(b) In order to increase the pH to 5.00, we need to calculate the new concentration of H^+ ions:
pH = -log10[H^+]
5.00 = -log10[H^+]
[H^+] = 10^(-5) = 1.0 x 10^(-5) M
We can use the expression for Ka again to find the new concentrations of HA and A^-:
Ka = (H^+)(A^-)/(HA)
We know that the amount of NaOH added will react with the original HA present. Let x moles of NaOH be added to the solution:
HA + NaOH ==> NaA + HOH
0.12 - x mol x mol
Now we can use the updated concentrations of HA and A^- in the Ka expression:
Ka = (1 x 10^(-5))(0.08 + x)/(0.12 - x)
We know the Ka value:
1.05 x 10^(-5) = (1 x 10^(-5))(0.08 + x)/(0.12 - x)
Solve for x:
(1.05)(0.08 + x) = (1)(0.12 - x)
0.084 + 1.05x = 0.12 - x
2.05x = 0.036
x = 0.036 / 2.05 = 0.0176
So, 0.0176 moles of NaOH should be added to the solution to increase the pH to 5.00.
A hypothetical weak acid HA, was combined with NaOH in the following proportions: 0.20 mol HA, 0.08 mol NaOH. The mixture was then diluted to a total volume of 1L, and the pH measured.
(a) If pH=4.80, what is the pKa of the acid
(b) how many additional moles of NaOH should be added to the solution to increase the pH to 5.00
HA + NaOH ==> NaA + HOH
0.2 mol HA to begin.
0.08 mol NaOH used to neutralize.
0.2-0.08 = 0.12 mol HA reamins.
0.08 mol A^- formed.
M = mols/L.
Ka = (H^+)(A^-)/(HA)
You know (H^+), from pH, you know A^- and you know HA. Solve for Ka, then pKa. Post your work if you get stuck.
1 answer