Question

(1 pts) Concentration of standard HCl solution (M) Table view List view Trial 1 Trial 2 Initial burette reading (ml) Final burette reading (mL) Volume of HCl added (mL) Solution temperature (°C)

Answer 1

Concentration of standard HCl solution(M) is 0.050 M

Trial 1 Trial 2

Initial burette reading(mL) 1.3 1.7

Final burette reading (mL) 12.5 13.1

Volume of HCl added (mL) 11.2 11.4

Solution temperature (⁰C) 25 25

Average volume HCl added(mL) = (11.2+11.4)/2 = 11.3

Concentration of OH^- (M) = 5.65 x 10^-4/ 0.02 = 0.02825 [Here in this question the initial volume of Ca(OH)₂ is not given, so as per standard procedure it is taken as 20 mL.]

Calculation: number of moles of H⁺= Number of moles of OH^-

Number of moles of H⁺ = 11.3 X 0.050/1000 = 5.65 x 10^-4

Concentration of Ca²⁺ = Concentration of OH^- (M)/2 = 0.014125

Calculation: In Ca(OH)₂ , Ca²⁺: OH^- = 1:2

Value of K_sp for  Ca(OH)₂  = [Ca²⁺] x [OH^-]² = 0.014125 X (0.02825)² = 1.12 x 10^-5 M³

Concentration of diluted HCl solution is 3.8856 x 10^-3 M

Calculation, M1= 0.09714 M, V1= 10 mL, V2= 250 mL

So, M2= M1V1/V2= 0.09714 X 10/250 = 3.8856 X 10^-3 M

If, 7.93 mL of diluted HCl is required to reach the endpoint, then concentration of OH^- = 1.232 x 10^-4 M

Calculation:

Number of moles of H⁺= Number of moles of OH^-

Number of moles of H⁺ = 7.93 x 3.8856 x 10^-3/1000 = 3.08 x 10^-5

Concentration of OH^- = 3.08 x 10^-5/0.25 = 1.232 X 10^-4 M

Concentration of Ag⁺ in the solution is 1.232 X 10^-4 M

Calculation: In AgOH, Ag⁺:OH^-= 1:1

K_sp(AgOH) = [Ag⁺][OH^-]

K_sp(AgOH) = (1.232 X 10^-4)² = 1.517 X 10^-8 M²