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.
(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.
To find the pKa of the weak acid HA, we need to find the concentration of H+ ions, A- ions, and HA in the solution.
Given that the pH of the solution is 4.80, we know that [H+] = 10^(-pH) = 10^(-4.80) = 1.58 x 10^(-5) M.
Since 0.20 mol of HA was initially added and 0.08 mol of NaOH was used to neutralize it, we have 0.12 mol of HA remaining.
The reaction between HA and NaOH is as follows:
HA + NaOH -> NaA + H2O
Since the weak acid is monoprotic, 1 mole of HA reacts with 1 mole of NaOH to produce 1 mole of A- ions. Therefore, we have 0.08 mol of A- ions in the solution.
Now we can calculate the concentration of HA:
[HA] = moles/volume = 0.12 mol / 1 L = 0.12 M
And the concentration of A- ions:
[A-] = 0.08 mol / 1 L = 0.08 M
Using the equation for the dissociation of the weak acid:
Ka = ([H+][A-]) / [HA]
Substituting the values we have:
Ka = (1.58 x 10^(-5) M)(0.08 M) / 0.12 M = 1.05 x 10^(-5)
To find the pKa, we take the negative logarithm of Ka:
pKa = -log(Ka) = -log(1.05 x 10^(-5)) = 4.98
Therefore, the pKa of the weak acid HA is approximately 4.98.
Now, let's move on to part (b) of the question.
To increase the pH from 4.80 to 5.00, we need to calculate the additional moles of NaOH required.
We can use the Henderson-Hasselbalch equation to relate the pH, pKa, and the ratio of concentrations of A- ions to HA:
pH = pKa + log([A-] / [HA])
Since we are aiming for a pH of 5.00, we can rearrange the equation to solve for the ratio of [A-] to [HA]:
log([A-] / [HA]) = pH - pKa
[A-] / [HA] = 10^(pH - pKa)
Substituting the known values:
[A-] / [HA] = 10^(5.00 - 4.98) = 10^0.02 ≈ 1.02
This means the concentration of A- ions is approximately 1.02 times the concentration of HA.
Since we currently have 0.08 mol of NaOH in the solution (which corresponds to 0.08 mol of A-), the additional moles of NaOH required can be calculated as follows:
Additional moles of NaOH = (A- / HA) * additional moles of HA
Additional moles of NaOH = (1.02 * 0.12) mol ≈ 0.1224 mol
Therefore, approximately 0.1224 moles of NaOH should be added to the solution to increase the pH from 4.80 to 5.00.
To find the pKa of the acid (HA), let's go step by step:
(a) First, let's find the concentration of HA (initial and remaining):
Initial concentration of HA = 0.20 mol/1L = 0.20 M
Since 0.08 mol of NaOH was used to neutralize the acid, the remaining concentration of HA can be found by subtracting the moles of NaOH used from the initial concentration of HA:
Remaining concentration of HA = Initial concentration of HA - Moles of NaOH used
Remaining concentration of HA = 0.20 M - 0.08 mol/1L = 0.12 M
(b) Next, let's find the concentration of A^- (the conjugate base of HA):
The moles of A^- formed from the reaction between HA and NaOH is equal to the moles of NaOH used:
Concentration of A^- = 0.08 mol/1L = 0.08 M
(c) Now, we can calculate the Ka of the acid using the equation: Ka = [H+][A^-]/[HA]
We can assume that all of the HA dissociates into H+ and A^- ions in solution. So, the concentration of H+ is equal to 10^(-pH):
[H+] = 10^(-pH) = 10^(-4.80) = 1.58 x 10^(-5) M
Using the given concentrations of A^- and HA, we can solve for Ka:
Ka = [H+][A^-]/[HA] = (1.58 x 10^(-5)) * (0.08)/(0.12) = 1.05 x 10^(-5) M
(d) Finally, to find the pKa, we use the equation: pKa = -log(Ka)
pKa = -log(1.05 x 10^(-5)) ≈ 4.98
Therefore, the pKa of the acid (HA) is approximately 4.98.
To answer part (b) of the question, let's see how many additional moles of NaOH are required to increase the pH from 4.80 to 5.00:
The change in pH is 5.00 - 4.80 = 0.20
To calculate the concentration of H+ at pH 5.00, we use the equation: [H+] = 10^(-pH) = 10^(-5.00) ≈ 1.00 x 10^(-5) M
To neutralize the additional H+ ions and increase the pH to 5.00, we need to add moles of NaOH:
Moles of NaOH needed = Additional H+ ions / (Concentration of NaOH)
Moles of NaOH needed = (1.00 x 10^(-5) M) / (0.08 M) = 1.25 x 10^(-4) mol
Therefore, approximately 1.25 x 10^(-4) moles of NaOH need to be added to the solution to increase the pH from 4.80 to 5.00.