What volume of 0.0100 M NaOH must be added to 1.00 L of 0.0500 M HOCl to achieve a pH of 8.00? K_a for HOCl is 3.5 × 10^-8. A) 1.2 L B) 1.0 L C) 3.9 L D) 5.0 L E) none of these

Question

Quizplus · Accepted Answer

Step 1: Write the balanced chemical equation for the reaction between HOCl and NaOH:
HOCl + NaOH ➞ NaOCl + H2O

Step 2: Write the expression for the equilibrium constant (Kb) for the reaction:
Kb = [NaOCl][H2O]/[HOCl][NaOH]

Step 3: Write the expression for the pH of the solution:
pH = pKa + log([NaOCl]/[HOCl])

Step 4: Substitute the values given in the problem:
pH = 8, Ka = 3.5 × 10^-8, [HOCl] = 0.0500 M

Solve for [NaOCl]:
8.00 = -log(3.5 × 10^-8) + log([NaOCl]/0.0500)
[NaOCl]/0.0500 = 10^-8.00 × 3.5 × 10^-8
[NaOCl] = (10^-8.00 × 3.5 × 10^-8) × 0.0500 = 1.75 × 10^-15 M

Step 5: Calculate the volume of NaOH needed to reach the desired pH:
Assuming all of the NaOH reacts with HOCl and none remains in excess, the amount of NaOH needed can be calculated using the stoichiometry of the balanced equation:
moles of NaOH = moles of HOCl = [HOCl] × VHOCl
moles of NaOH = [NaOCl]/2 = 1.75 × 10^-15/2 = 8.75 × 10^-16 mol
VNaOH = moles of NaOH / [NaOH] = 8.75 × 10^-16 mol / 0.0100 M = 8.75 × 10^-14 L = 3.9 L

Therefore, the correct answer is (C) 3.9 L.

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