Question

im not sure if run 1 is correct. can someone help me with run 2? thanks

Name: Partner(s) Name Date: Part A: Molar Solubility and Ksp of Calcium Hydroxide Molarity of standardized HCl solution (mol/L): 0.050 44 RUN 1 RUN 2 Initial buret reading (mL) 0.00 0.00 Volume of HCl added at equivalence point (mL) 192 19.4 Volume of HCI added at equivalence point (L) og 10.194 10.875 mol 10-05) Moles of HCI added at equivalence point Moles of OH in saturated Ca(OH)2 solution 8.75X104 Volume of saturated Ca(OH)2 solution (L) 0.01006 3.61606 Molarity of OH at equilibrium, (OH) (1/2)[0H Valo0205m Molarity of Ca²+ at equilibrium, (Ca2+1 Molar solubility of Ca(OH)2 0-0205m O oloasm 1901x10 3 4.3x10 6 Kap of Ca(OH)2 (coot] [oH =00ROSE Average molar solubility of Ca(OH)2 0024 Average Ksp of Ca(OH)2 Show detailed calculation of molar solubility and Ksp for Run 1:

Answer 1

For Run 1,

Volume of HCl added at equivalence point = 19.1 mL

Volume of HCl added at equivalence point = 19.1 mL * (1 L / 1000 mL)

Volume of HCl added at equivalence point = 0.0191 L

moles HCl added at equivalence point = (molarity HCl) * (Volume of HCl added at equivalence point in Liter)

moles HCl added at equivalence point = (0.05044 M) * (0.0191 L)

moles HCl added at equivalence point = 9.634 x 10^-4 mol

moles OH^- = moles HCl added at equivalence point

moles OH^- = 9.634 x 10^-4 mol

volume of Ca(OH)₂ solution = 25 mL = 0.025 L

molarity OH^- at equilibrium, [OH^-] = (moles OH^-) / (volume of Ca(OH)₂ solution in Liter)

[OH^-] = (9.634 x 10^-4 mol) / (0.025 L)

[OH^-] = 0.03854 M

molarity of Ca²⁺ at equilibrium, [Ca²⁺] = [OH^-]/2

[Ca²⁺] = (0.03854 M) / 2

[Ca²⁺] = 0.01926 M

molar solubility of Ca(OH)₂ = [Ca²⁺]

molar solubility of Ca(OH)₂ = 0.01926 M

K_sp Ca(OH)₂ = [Ca²⁺][OH^-]²

K_sp Ca(OH)₂ = (0.01926 M) * (0.03854 M)²

K_sp Ca(OH)₂ = 2.86 x 10^-5

Similarly for Run 2

Volume of HCl added at equivalence point = 0.0194 L

moles of HCl added at equivalence point = 9.785 x 10^-4 mol

mole of OH^- in saturated Ca(OH)₂ solution = 9.785 x 10^-4 mol

volume of saturated Ca(OH)₂ solution = 0.025 L

[OH^-] = 0.03914 M

[Ca²⁺] = 0.01957 M

molar solubility of Ca(OH)₂ = 0.01957 M

K_sp Ca(OH)₂ = 3.0 x 10^-5

Average K_sp of Ca(OH)₂ = (K_sp Ca(OH)₂ for Run 1 + K_sp Ca(OH)₂ for Run 2) / 2

Average K_sp of Ca(OH)₂ = (2.86 x 10^-5 + 3.0 x 10^-5) / 2

Average K_sp of Ca(OH)₂ = (5.86 x 10^-5) / 2

Average K_sp of Ca(OH)₂ = 2.93 x 10^-5