You add an excess of Ca(OH)_2 to water maintained at a particular temperature, s

Question

You add an excess of Ca(OH)_2 to water maintained at a particular temperature, stir until the solution is saturated, filter, then determine the [OH^-] in the solution by titration with acid. Titration of 10.0 mL of the calcium hydroxide solution to the end point requires 4.86 mL of 0.070 M HCl solution. What is the molar quantity of OH^- in the 10.0 mL of solution? What are the concentrations of Ca^2+ and OH? What is the solubility of Ca(OH)_2 under these conditions? What is the K_sp for Ca(OH)_2 under these conditions? You measure the solubility of a salt at four different temperatures and calculate the following K_sp values: delta G degree _soln for the solubility process is given by the relationship delta G degree _soln = -RT ln K_sp where R = 8.314 J/(K times mol), and T is the absolute temperature in degrees K. Calculate delta G degree _soln (in kJ/mol) at each of these temperatures. As the temperature increases, how does the solubility of the salt change? According to lessthanorequalto Chatelier's principle, dissolving this salt should therefore be which type of process? In other words, what is the sign of delta H degree_soln?

Explanation / Answer

1) calculation of molar quantity of OH-

V1=volume of OH-=10ml

V2=volume of acid required in titration=4.86ml

M1=molarity of OH-

M2=molarity of acid=0.070M

M1V1=M2V2=moles of acid/base

So moles of OH- in the solution =M2V2=0.070 mol/L*4.86 ml*(1L/1000ml)=0.000342 moles

2) concentration of OH- =molarity of OH-=M1=M2V2/V1=0.000342moles/10ml=0.000342moles/0.01L=3.42*10^-6M

Ca(OH)2<--->Ca2+ +2OH-

[Ca2+]=1/2 [OH-]=1/2*3.42*10^-6M=1.71*10^-6M

concentration of OH-=1.71*10^-6M

3)solubility ofCa(OH)2=[Ca2+]=1.71*10^-6M

4)ksp=[Ca2+] [OH-]^2=(1.71*10^-6M)(3.42*10^-6M)^2=5.85*10^-18 M^3

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