A 20.2 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 5.50. What is the Ka of HX?
millimols HX initially = 50 x 0.06 = 3.00
mmols NaOH added for pH 5.50 = 30.0 x 0.06 = 1.8
............HX + NaOH ==> NaX + H2O
initial....3.00....0.......0......0
add...............1.8...............
change.....-1.8..-1.8.......+1.8...+1.8
equil.......1.2.....0......1.8......1.8
Plug 5.50 for pH and the other info into the Henderson-Hasselbalch equation. Solve for pKa and convert top Ka.
6.5
To find the Ka of HX, we need to understand the chemical reaction that occurs between HX and NaOH. HX is a weak monoprotic acid, which means it donates only one proton (H+) in a chemical reaction. NaOH is a strong base and reacts with HX to form water and the salt sodium salt (NaX).
The chemical equation for the reaction between HX and NaOH is:
HX + NaOH -> H2O + NaX
To solve this problem, we can use the concept of stoichiometry. Since we know the volume and molarity of NaOH needed to reach the equivalence point, we can calculate the number of moles of NaOH that reacted.
Number of moles of NaOH = volume of NaOH (in L) × molarity of NaOH
= 50.0 mL × (1 L/1000 mL) × 0.060 M
= 0.003 moles
Since HX and NaOH react in a 1:1 mole ratio according to the balanced equation, we can conclude that the number of moles of HX is also 0.003 moles.
Now, let's calculate the initial concentration of HX:
Initial concentration of HX = number of moles of HX / volume of HX solution (in L)
= 0.003 moles / (20.2 mL × (1 L/1000 mL))
= 0.148 M
At the equivalence point, all the HX has reacted with NaOH, resulting in the formation of NaX. This means the remaining volume of NaOH added (30.0 mL) reacted with 0.003 moles of HX.
Now, let's calculate the concentration of NaOH at the equivalence point:
Concentration of NaOH at equivalence point = number of moles of NaOH / volume of NaOH added after pH measurement (in L)
= 0.003 moles / (30.0 mL × (1 L/1000 mL))
= 0.100 M
Since NaOH is a strong base, it completely dissociates in water, giving an equal concentration of OH- ions. So, at the equivalence point, the concentration of OH- ions is also 0.100 M.
From the pH value provided (pH = 5.50), we can calculate the concentration of H+ ions. The pH is given by the equation:
pH = -log[H+]
Therefore,
[H+] = 10^(-pH)
[H+] = 10^(-5.50)
= 3.16 × 10^(-6) M
Since HX is a weak acid, it partially dissociates in water. Assuming x is the concentration of the H+ ions at the equilibrium, we can set up the equilibrium expression for the dissociation of HX as:
HX ⇌ H+ + X-
Ka = [H+][X-] / [HX]
We know the concentration of H+ ions is 3.16 × 10^(-6) M, and since HX and NaX react in a 1:1 ratio, the concentration of X- ions is also 3.16 × 10^(-6) M.
Plugging these values into the equilibrium expression:
Ka = (3.16 × 10^(-6) M) × (3.16 × 10^(-6) M) / (0.148 M)
Ka = 6.77 × 10^(-9)
Therefore, the Ka of HX is 6.77 × 10^(-9).