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During the titration of 100 ml of a weak monobasic acid solution using 0.1 M NaOH, the solution became neutral at 40 mL addition of NaOH and equivalence point was obtained at 50 mL NaOH addition. The Ka of the acid is (log 2 = 0.3)
  • a)
    3 × 10–7
  • b)
    4 × 10–7
  • c)
    1 × 10–7
  • d)
    2 × 10–7
Correct answer is option 'B'. Can you explain this answer?
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First, let's determine the moles of NaOH added at the equivalence point.

Moles of NaOH = concentration of NaOH * volume of NaOH added
Moles of NaOH = 0.1 M * 50 mL
Moles of NaOH = 0.1 mol/L * 0.05 L
Moles of NaOH = 0.005 mol

Since the weak monobasic acid is a monoprotic acid, the moles of acid at the equivalence point will be the same as the moles of NaOH added.

Moles of acid = 0.005 mol

Now, let's determine the initial moles of the weak monobasic acid.

Initial moles of acid = concentration of acid * volume of acid
Initial moles of acid = unknown concentration * 0.1 L

Since we don't know the concentration of the acid, let's call it "C".

Initial moles of acid = C * 0.1 mol

At the halfway point (when the solution becomes neutral), half of the initial moles of acid will have reacted with half of the moles of NaOH added.

Moles of acid at halfway point = 0.5 * initial moles of acid
Moles of acid at halfway point = 0.5 * C * 0.1 mol

Since the solution becomes neutral at the halfway point, the concentration of the conjugate base will be equal to the concentration of the acid.

Therefore, the concentration of the conjugate base at the halfway point is also 0.5 * C mol/L.

The Ka expression for the acid is: Ka = [conjugate base] * [H3O+] / [acid]
At the halfway point, the concentration of the conjugate base is 0.5 * C mol/L, and the concentration of the acid is also 0.5 * C mol/L.
The concentration of H3O+ at the halfway point is unknown, so let's call it "x".

Ka = (0.5 * C) * x / (0.5 * C)

Since the solution is neutral at the halfway point, the concentration of H3O+ is equal to the concentration of OH- at the equivalence point, which is 0.005 mol/L.

Ka = (0.5 * C) * 0.005 mol/L / (0.5 * C)

The C cancels out:

Ka = 0.005 mol/L

Since log 2 = 0.3, we can rearrange the equation to solve for Ka:

Ka = 10^(0.3)
Ka ≈ 2

Therefore, the Ka of the acid is approximately 2.
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During the titration of 100 ml of a weak monobasic acid solution using 0.1 M NaOH, the solution became neutral at 40 mL addition of NaOH and equivalence point was obtained at 50 mL NaOH addition. The Kaof the acid is (log 2 = 0.3)a)3 × 10–7b)4 × 10–7c)1 × 10–7d)2 × 10–7Correct answer is option 'B'. Can you explain this answer?
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