Question

In: Chemistry

a) A student A weighed .50 Ca(OH)2 for his saturated solution and student B weighed 0.23...

a) A student A weighed .50 Ca(OH)2 for his saturated solution and student B weighed 0.23 g, Both students dissolved it in 60 ml of water and both had solid left in their solution before filtrating. After filtrating to a clear solution, which student had the most concentrated solution? What would the effect on the volume of HCl titrant used and the calculated Ksp of the reaction if one of the students had a couldy solution after filtering and proceeded to titrate it compared to a clear solution?

(b) A saturated solution of CA(OH)2 is prepared and filtered. If 9.32 mL of 0.045 M HCl is used to titrate a 10 ml sample of the solution to end point:
- how many moles of Ca^2+, OH- were in the original 10 mL sample
- what are the concentrations of Ca^2+ and OH- in the saturated solution?
- what is the molar solubility for Ca(OH)2 under these conditions?
- Calculate Ksp for Ca(OH)2 under these conditions

(c) If change in entropy of the universe can tell us whether a process will be spontaneous, why do we need delta G (Gibb's free energy)

Solutions

Expert Solution

Answer – A) We are given, student A weighed .50 Ca(OH)2 for his saturated solution and student B weighed 0.23 g. volume of Ca(OH)­2 dissolve from both student = 60 mL

We know each compound has specific solubility for each 100 mL of solution. We know Ca(OH)2 is slightly soluble in water, so in the 60 mL of water Ca(OH)2 dissolved same amount from both student, since solubility for Ca(OH)2 at 60 mL water is fix and remaining ca(OH)2 gets saturated and not dissolved. So both student has the same concentration.

b) We are given, solution of Ca(OH)2 reacts with HCl solution.

[HCl] = 0.045 M , volume = 9.32 mL , volume of Ca(OH)2 solution = 10 mL

We know the reaction

Ca(OH)2 + 2 HCl ----> CaCl2 + 2 H2O

Now we need to calculate the moles of HCl

Moles of HCl = 0.045 M * 0.00932 L

                       = 0.000419 moles

We know

2 moles of HCl = 1 moles of Ca(OH)2

So, 0.000419 moles of HCl = ?

= 0.00021 moles of Ca(OH)2

Now moles of Ca2+ moles were in the original 10 mL sample

1 moles of Ca(OH)2 = 1 moles of Ca2+

So, 0.00021 moles of Ca(OH)2 = ?

= 0.00021 moles of Ca2+.

Now moles of OH- in were in the original 10 mL sample

1 moles of Ca(OH)2 = 2 moles of OH-

So, 0.00021 moles of Ca(OH)2 = ?

= 0.000419 moles of OH-

-concentrations of Ca2+ and OH- in the saturated solution –

[Ca2+] = 0.00021 moles / 0.010 L

            = 0.021 M

[OH-] = 0.000419 moles / 0.010 L

            = 0.042 M

Ksp for Ca(OH)2

We know Ksp expression for the Ca(OH)2

Ksp = [Ca2+] [OH-]2

        = (0.021)*(0.042)2

         = 3.7*10-5

c) Change in entropy of the universe can tell us whether a process will be spontaneous, but we need delta G, because for the same interconverting like solid to solid phases and small change not counted by entropy, but free energy gives us the same interconverting like solid to solid phases and small change. When free energy is negative then spontaneous process.


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