A 22.6 mL sample of a weak monoprotic acid, HX, requires 50.0 mL of 0.050 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?

Surely you have made a typo (or two). There is no way the pH can be 4.90. The equivalence point for the titration of a weak acid and strong base is approximately 8.3 or so. Adding BASE to that will make it higher, not lower.

To find the Ka of HX, we need to use the information provided about the volume and concentration of the weak acid and strong base.

Let's break down the steps to get the answer:

1. Calculate the number of moles of NaOH used to reach the equivalence point:
Moles of NaOH = Volume of NaOH × Concentration of NaOH
Moles of NaOH = 50.0 mL × 0.050 M
Moles of NaOH = 0.050 mol

2. Since NaOH is a strong base and reacts in a 1:1 ratio with the weak acid, the number of moles of NaOH is equal to the number of moles of HX:
Moles of HX = 0.050 mol

3. Now we need to calculate the initial concentration (M) of HX using the volume and moles:
Initial Concentration of HX = Moles of HX / Initial Volume of HX
Initial Concentration of HX = 0.050 mol / 22.6 mL

However, before calculating the concentration, we need to convert mL to L:
Initial Concentration of HX = 0.050 mol / (22.6 mL × 10^(-3) L/mL)
Initial Concentration of HX = 2.21 M

4. The pH after the addition of 30.0 mL of NaOH and before reaching the equivalence point can be used to calculate the pKa (the negative logarithm of Ka):
pH = pKa + log10 (Initial Concentration of HX / Remaining Concentration of HX)

Rearranging the equation and substituting the values:
pKa = pH - log10 (Initial Concentration of HX / Remaining Concentration of HX)
pKa = 4.90 - log10 (2.21 / X)

Where X is the remaining concentration of HX.

5. Finally, we can use the pKa to calculate the Ka value. Ka is the antilog of -pKa:
Ka = 10^(-pKa)

Now, we can substitute the values into the equation and solve for Ka.