An HCl solution has a concentration of 0.09714 M. Then 10.00 mL of this solution was then diluted to 250.00 mL in a volumetric flask. The diluted solution was then used to titrate 250.0 mL of a saturated AgOH solution using methyl orange indicator to reach the endpoint. (1pts) 1. What is the concentration of the diluted HCl solution? (1pts) 2. If 7.93 mL of the diluted HCl solution was required to reach the endpoint, what is the concentration of OH− in solution? (1pts) 3. What is the concentration of Ag+ in solution? (1pts) 4. What is the Ksp expression for the dissolution of AgOH? Ksp = [Ag+] + [OH−] Ksp = [Ag+] + [OH−]2 Ksp = [Ag+][OH−]2 Ksp = [Ag+][OH−] Choose... (1pts) 5. What is the value of the Ksp of AgOH?

1. 0.003886M HCl

2. 1.233x10⁻⁴M OH⁻

3. 1.233x10⁻⁴M Ag⁺

4. Ksp = [Ag⁺] [OH⁻]

5. Ksp = 1.519x10⁻⁸

Explanation:

1. As the HCl solution was diluted from 10.00mL to 250.00mL, the dilution factor is:

250.00mL / 10.00mL = 25

That means the HCl solution was diluted 25 times

The undiluted solutio is 0.09714M HCl, and diluted solution is:

0.09714M / 25 = 0.003886M HCl

2. The reaction of HCl with OH⁻ ions is:

HCl + OH⁻ → H₂O + Cl⁻

Where 1 mole of HCl reacts per mole of OH⁻

As 7.93mL of the solution = 7.93x10⁻³L of HCl solution are used to titrate, moles are:

7.93x10⁻³L ₓ (0.003886mol / L) = 3.081x10⁻⁵ moles of HCl are used in the titration and that means there are 3.081x10⁻⁵ moles of OH⁻ are in solution.

As the AgOH solutioin was 250.0mL = 0.2500L, molar concentration of OH⁻ is:

3.081x10⁻⁵ moles of OH⁻ / 0.2500L = 1.233x10⁻⁴M OH⁻

3. In the equilibrium of AgOH in water:

AgOH(s) ⇄ Ag⁺(aq) + OH⁻(aq)

Moles of Ag⁺ in solution are equal to moles of OH⁻, that is 1.233x10⁻⁴M Ag⁺ is the concentration of Ag⁺ ions.

4. Again, in the equilibrium:

AgOH(s) ⇄ Ag⁺(aq) + OH⁻(aq)

Ksp is defined as:

Ksp = [Ag⁺] [OH⁻]

Because concentrations of solids doesn't affect the equilibrium

5. Replacing in Ksp formula:

Ksp = [1.233x10⁻⁴M OH⁻] [1.233x10⁻⁴M OH⁻]

Ksp = 1.519x10⁻⁸

your answer would be a. 12.5 kg