While titrating 25.00mL of a weak acid, HA, with 0.1500M NaOH, you reach equivalence point after adding 27.00mL of the NaOH. The pH of the acid initially was 2.48. What is the dissociation constant of the acid?
HA ==> H^+ + A^-
K = (H^+)(A^-)/(HA) = Ka
If you know NaOH molarity and NaOH volume, can you calculate the molarity of HA initially? Then knowing (HA), use pH initially to determine (H^+). Of course, (A^-) will be the same (you know this from the ionization equation above. That leaves just Ka to calculate. Post your work if you get stuck.
Correction of one more typo:
[H+] = [A-] = [HA]
should change to:
[H+] = [A-] =3.31x10^-3 M
To find the dissociation constant of the weak acid, HA, we need to use the equation that relates the initial pH, the volume of the weak acid, the volume of the strong base, and the dissociation constant.
The equation we use is:
pH = pKa + log([A-]/[HA])
Where:
- pH is the initial pH of the weak acid
- pKa is the dissociation constant of the weak acid
- [A-] is the concentration of the conjugate base of the weak acid
- [HA] is the concentration of the weak acid
Let's calculate the concentration of the weak acid, [HA]:
Given that the volume of the weak acid is 25.00 mL, and assuming its density is equal to that of water, we can convert this to liters:
Volume of HA = 25.00 mL = 0.02500 L
Now we can calculate the concentration of the weak acid, [HA]:
[HA] = moles of HA / volume of HA
To find the moles of HA, we can use the equation:
moles of HA = concentration of HA × volume of HA
Given that the concentration of the NaOH is 0.1500 M and the volume of NaOH added is 27.00 mL, we can calculate the moles of NaOH:
moles of NaOH = concentration of NaOH × volume of NaOH
moles of NaOH = 0.1500 M × 0.02700 L
moles of NaOH = 0.00405 mol
Since NaOH reacts with HA in a 1:1 ratio, the moles of HA are also 0.00405 mol.
Now we can calculate the concentration of the weak acid, [HA]:
[HA] = 0.00405 mol / 0.02500 L
[HA] = 0.162 M
Now that we have the concentration of the weak acid, [HA], we can rearrange the equation to solve for the dissociation constant, pKa:
pKa = pH - log([A-]/[HA])
Given that the pH is 2.48 and the volume of NaOH added is equal to the volume at the equivalence point, we can find the concentration of the conjugate base, [A-]:
The volume of NaOH added at the equivalence point is 27.00 mL, which we can convert to liters:
Volume of NaOH at equivalence point = 27.00 mL = 0.02700 L
The moles of NaOH at equivalence point can be calculated using the same equation as before:
moles of NaOH = concentration of NaOH × volume of NaOH
moles of NaOH = 0.1500 M × 0.02700 L
moles of NaOH = 0.00405 mol
Since NaOH reacts with HA in a 1:1 ratio, the moles of the conjugate base, A-, are also 0.00405 mol.
Now let's calculate the concentration of the conjugate base, [A-]:
[A-] = moles of A- / volume of HA
[A-] = 0.00405 mol / 0.02500 L
[A-] = 0.162 M
Now we can substitute the values into the equation to find the dissociation constant, pKa:
pKa = 2.48 - log(0.162/0.162)
pKa = 2.48 - log(1)
Since log(1) = 0, the equation simplifies to:
pKa = 2.48 - 0
Therefore, the dissociation constant, pKa, is equal to 2.48.
The dissociation constant of the weak acid, HA, is the antilog of pKa:
Ka = 10^(-pKa)
Ka = 10^(-2.48)
Using a calculator, the dissociation constant, Ka, is approximately 3.011 x 10^(-3).
The titration helps determine the molar concentration of HA:
Moles of HA(aq) = moles of OH- used
moles of OH- = (0.1500Mmol./L)(0.02500L) = 4.050x10^-3 = mol. HA
4.050x10^-3 mol. HA / 0.02500L = 0.162 M HA
Next we find the H+ ion concentration
[H+] = 10^-pH = 10^-2.48 = 3.31x10^-3 M
Finally we find the Ka:
Ka = [H+][A-] / [HA]
[H+] = [A-] = [HA]
[HA] = 0.162 M - 3.31x10^-3 M
Substitute into the Ka expression to get the value of Ka.