A 125.0 mg sample of an unknown, monoprotic acid was dissolved in 100.0 mL of distilled water and
titrated with a 0.050 M solution of NaOH. The pH of the solution was monitored throughout the titration,
and the following data were collected. Determine the Ka of the acid.
Volume of
OH- (mL) 0 5 10 15 17 18 19 20 21 22 22.2 22.6 22.8 23 24 25 30 40
pH 3.09 3.65 4.10 4.50 4.55 4.71 4.94 5.11 5.37 5.93 6.24 9.91 10.2 10.4 10.8 11.0 11.5 11.8
To determine the Ka of the acid, we first need to find the equivalence point of the titration. The equivalence point is the volume of NaOH solution added where the number of moles of acid is equal to the number of moles of base.
Let's start by finding the equivalence point volume:
1. Locate the data point where the pH starts to rise sharply. This indicates that we are close to the equivalence point.
- From the given data, we can see the pH value rises sharply between 22.2 mL and 22.6 mL of NaOH solution.
2. Use linear interpolation to estimate the exact volume of NaOH solution at the equivalence point.
- Let's calculate the halfway point between 22.2 mL and 22.6 mL:
Equivalence point volume = (22.2 + 22.6) / 2 = 22.4 mL
Now that we have the equivalence point volume, we can proceed to calculate the concentration of the acid at the equivalence point, which will be used to determine the Ka value.
3. Calculate the number of moles of NaOH at the equivalence point.
- Volume of NaOH solution at the equivalence point = 22.4 mL = 0.0224 L
- Moles of NaOH = concentration of NaOH (M) x volume of NaOH solution (L)
Moles of NaOH = 0.050 M x 0.0224 L = 0.00112 mol
4. Use the stoichiometry of the balanced chemical equation to determine the number of moles of acid at the equivalence point.
- Since the acid is monoprotic, the balanced chemical equation is:
Acid + NaOH → Salt + Water
This means that the ratio between the moles of acid and NaOH is 1:1.
Therefore, the moles of acid at the equivalence point = 0.00112 mol
5. Calculate the concentration of the acid at the equivalence point.
- Volume of water used = 100.0 mL = 0.1000 L
- Concentration of acid at the equivalence point = moles of acid / volume of water used
Concentration of acid at the equivalence point = 0.00112 mol / 0.1000 L = 0.0112 M
Now, we can determine the Ka of the acid using the pH value at the half-equivalence point.
6. Locate the data point where the pH is halfway between the initial pH and the pH at the equivalence point.
- From the given data, we see that the halfway point between the initial pH (3.09) and the pH at the equivalence point (6.24) is between 22 and 22.2 mL of NaOH solution.
7. Use linear interpolation to estimate the exact volume of NaOH solution at the half-equivalence point.
- Let's calculate the halfway point between 22 mL and 22.2 mL:
Half-equivalence point volume = (22 + 22.2) / 2 = 22.1 mL
8. Find the corresponding pH value at the half-equivalence point.
- From the data, at 22.1 mL of NaOH solution, the pH is 6.24.
Now, we can calculate the pKa of the acid using the pH value at the half-equivalence point.
9. Calculate pKa using the Henderson-Hasselbalch equation:
pH = pKa + log([A-] / [HA])
Since the acid is monoprotic, [A-] = concentration of acid at the equivalence point.
Let's denote the pKa as x for simplicity.
6.24 = x + log(0.0112 / 0.0112)
log(1) = 6.24 - x
0 = 6.24 - x
10. Solve for x (pKa):
x = 6.24
Therefore, the pKa of the acid is 6.24.
Note: The Ka value can be obtained by taking the antilog of the pKa value:
Ka = 10^(-pKa)
Ka = 10^(-6.24)
To determine the Ka of the acid, we need to analyze the pH values at different volumes of NaOH added. The pH values will correspond to different stages of the titration, where the acid is neutralized by the base.
The titration begins with the addition of 0 mL of NaOH, indicating the initial pH of the acid before any base is added. From the given data, the initial pH is 3.09.
Plotting the titration curve (pH vs. volume of NaOH) using the collected data, we can observe the following:
At the start of the titration, before any base is added, the acid is present in its acidic form, and the pH is low.
As the volume of NaOH increases, the base reacts with the acid, neutralizing it. The pH gradually increases as the acid is converted to its conjugate base.
At the equivalence point (where the moles of acid and base are equal), the acid has been completely neutralized, and the pH is determined by the concentration of the conjugate base, in this case, the sodium salt of the acid. The pH at the equivalence point is indicated by a significant jump in pH, which in this case is around 10.8.
After the equivalence point, additional NaOH causes the pH to increase rapidly, indicating excess base presence in the solution.
To calculate the Ka of the acid, we need to find the volume of NaOH at half the equivalence point (half-neutralization) and use it to determine the concentration of the acid at that point. Then, we can apply the Henderson-Hasselbalch equation to calculate the Ka.
From the given data, we can see that the pH is 5.93 at a volume of 22 mL of NaOH. This volume is close to half the equivalence point, so we will assume it to be the half-neutralization volume.
Now, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Since the acid is monoprotic, [A-] (concentration of the conjugate base) is equal to the concentration of NaOH added (0.05 M) at the half-neutralization volume (22 mL). We can convert this to moles:
[A-] = (0.05 mol/L) * (0.022 L) = 0.0011 mol
To find [HA] (concentration of the acid), we need to calculate the moles of acid initially present in the solution. Since we have a 125.0 mg sample of the acid, we can convert this to moles:
[HA] = (125.0 mg) / (molar mass of the acid)
Let's assume the molar mass of the acid is M. Then:
[HA] = (125.0 mg) / M
Now, substituting the values into the Henderson-Hasselbalch equation:
5.93 = pKa + log(0.0011 mol / (125.0 mg / M))
Solving for pKa:
pKa = 5.93 - log(0.0011 mol / (125.0 mg / M))
Finally, we can calculate Ka:
Ka = 10^(-pKa)
Substitute the value of pKa we calculated into this formula to get the value of Ka.