A 20.1 mL sample of a weak monoprotic acid, HX, requires 50.0 mL of 0.060 M NaOH to reach the equivalence point. After the addition of 30.0 mL of NaOH, the pH is 4.90. What is the Ka of HX?
There are a couple of ways to do this; one is quite straight forward, the other requires a little reasoning power. Here is the straight forward way.
mLNaOH x MNaOH = mLHX x MHX
50.0 x 0.060 = 20.1 x MHX
Solve for Molarity HX. For simplicity I shall call this about 0.15M.
So millimiles HX to start = 20.1*0.15 = about 3.0
millimoles NaOH at 30 mL = 30 mL x 0.06 = 1.8 mmoles.
Set up an ICE chart.
...........HX + NaOH ==> NaX + H2O
initial....3.0...0........0......0
add.............1.8...............
react......-1.8..-1.8....1.8.....+1.8
equilibrium..1.2...0.....1.8......1.8
Substitute into Ka expression for HX (convert pH to H^+ first), and solve for Ka.
I worked it out to be about 1.9E-5 for Ka but check my work.
The more esoteric way of doing it is to use the Henderson-Hasselbalch equation.
You know, from the equivalence point, that if you add 3.0 millimoles NaOH that you must have started with 3.0 millimoles HX. So you add 1.8 millimoles NaOH at the 30 mL mark which means 1.2 HX is left and 1.8 mmoles were converted to X^-. Then substitute into pH = pKa + log (X^-)/(HX) and solve for pKa, then convert to Ka. I have 4.72 for pKa.
To determine the Ka of the weak monoprotic acid, HX, you can use the information provided.
First, let's analyze the titration process:
1. Initially, you have a 20.1 mL sample of HX.
2. It requires 50.0 mL of 0.060 M NaOH to reach the equivalence point.
3. After adding 30.0 mL of NaOH, the pH is 4.90.
The equivalence point occurs when the moles of HX react with an equal number of moles of NaOH. Therefore, we need to determine the moles of NaOH added at the equivalence point.
Moles of NaOH = volume (L) × concentration (M)
Moles of NaOH = 50.0 mL × 0.060 M = 0.003 moles
Since HX is a monoprotic acid, it reacts in a 1:1 ratio with NaOH. Therefore, the number of moles of HX at the equivalence point is also 0.003 moles.
Now, let's find the initial concentration of HX (before any NaOH is added):
Initial concentration of HX = moles of HX / volume (L)
Initial concentration of HX = 0.003 moles / (20.1 mL / 1000 mL/L) = 0.149 M
We can assume that the volume of the solution remains constant throughout the titration, so we can use the Henderson-Hasselbalch equation to find the pKa of HX at the halfway point (when 30.0 mL of NaOH has been added).
pH = pKa + log([A-]/[HA])
Since we are at the halfway point, the concentrations of HA and A- are equal.
pH = pKa + log(1)
4.90 = pKa
Therefore, the pKa of HX is 4.90.
To find the Ka, we need to convert the pKa to Ka:
Ka = 10^(-pKa)
Ka = 10^(-4.90)
Using a calculator, we find that the Ka of HX is approximately 7.94 × 10^(-5).