During the experiment, 25.00 mL of calcium hydroxide solution was titrated with

Question

During the experiment, 25.00 mL of calcium hydroxide solution was titrated with 5.50 mL of 0.102 M (a) Determine the mass of calcium hydroxide (in grams) used in the titration based on the data (b) Determine the solubility of the calcium hydroxide in 25.00 mL based on your answer to Part A. (c) Determine the Ksp for calcium hydroxide based on the data provided. (d) The literature value for the K_sp of calcium hydroxide is 7.90 times 10^-6. Calculate the % error for the data collected. The equation below can be used to determine the enthalpy and entropy change for the solubility of the calcium hydroxide. ln K = (- Delta H degree/R) 1/T + Delta S degree/R If a plot of ln K_sp vs 1/T produces a slope of +1, 875 K and a y-intercept of -18.5, calculate the values of both Delta H degree (in kJ/mol) and Delta S degree (in J/K-mol). Use your value of K from question 3d and R = 8.314 J/K-mol for the calculations.

Explanation / Answer

Q3.

V = 25 mL of Ca(OH)2 with HCl

ratio is

2HCl + Ca(OH)2 = CaCl2 + 2H2O

a)

mol of acid = MV = 5.5*0.102 = 0.561 mmol of acid

so... ratio is 1:1/2

mmol of base = 1/2 * mmol of acid = 1/2*0.561 = 0.2805 mmol of base

mass = mol*MW = (0.2805*10^-3)(74.093) = 0.02078 g of Ca(OH)2 = 20.78 mg of Ca(OH)2

b)

solubility of Ca(OH)2

[Ca(OH)2] = mmol / V = (0.2805)/(25) = 0.01122 mol of Ca(OH)2 per liter

c)

Ksp = [Ca+2][OH-]^2

so..

[Ca+2] = 0.01122 and [OH-] = 2*0.01122

Ksp = [Ca+2][OH-]^2 = (0.01122 )(2*0.01122 )^2 = 5.64987*10^-6

d)

Ksp = 7.9*10^-6

error% = (real - actual) / real * 100% = ((7.9*10^-6 ) - ( 5.64987*10^-6)) /  ( 7.9*10^-6 ) * 100 = 28.482% error

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