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? The K_a for HOCl is 3.5 $\times$10^-⁸.

Question

Quizplus · Accepted Answer

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

Step 2: Write the equation for the ionization of HOCl
HOCl + H2O ⇌ H3O+ + OCl-

Step 3: Write the expression for the acid dissociation constant (Ka) of HOCl
Ka = [H3O+][OCl-]/[HOCl]

Step 4: Write the expression for the pH of the solution
pH = -log[H3O+]

Step 5: Determine the equilibrium concentrations of HOCl, H3O+ and OCl- by using the ionization equation and the initial concentration of HOCl
Let x be the concentration of H3O+ and OCl-. Then, the concentration of HOCl is (0.0500 - x).
Ka = [H3O+][OCl-]/[HOCl]
3.5 × 10^-8 = x^2/(0.0500 - x)

Solving for x, we get x = 1.80 × 10^-4 M
Therefore, [H3O+] = [OCl-] = 1.80 × 10^-4 M and [HOCl] = 0.0482 M

Step 6: Determine the volume of NaOH required to raise the pH to 8.00
Since the pH is 8.00, [H3O+] = 1.00 × 10^-8 M
pH = -log[H3O+]
8.00 = -log(1.00 × 10^-8)
[H3O+] = 1.00 × 10^-8 M
[OH-] = Kw/[H3O+] = 1.00 × 10^-14/1.00 × 10^-8 = 1.00 × 10^-6 M

The amount of NaOH required to neutralize this amount of H3O+ is the same as the amount required to produce 1.00 × 10^-6 M OH- in solution.
Assuming NaOH is a strong base and completely dissociates, we can use the following equation to calculate the volume of NaOH required:
MNaOHVNaOH = MOH(M + + M -) × Vtot
0.0100 VNaOH = 1.00 × 10^-6 × 2.00 × Vtot
VNaOH = 2.00 × 10^-4/0.0100 = 0.0200 L
VNaOH = 20.0 mL

Converting mL to L, we get VNaOH = 0.0200 L or 20.0 mL. Therefore, the correct answer is D) 3.9 L is correct because it's 1.94 times more acidic than 10^-8 M H3O+ aqueous solution. For the pH to be increased to 8, the concentration of H+ ion should be decreased 100,000 times, which means that the new [H+] would be 1.00E-12 M. From this new concentration of [H+], one can obtain pOH as 4.0, and therefore pH would be 10.

What Volume of 0 ×\times× 10-8

What Volume of 0 $\times$ 10^-⁸