I did this chem experiment related to solubility and solubility product. had 4 f

Question

I did this chem experiment related to solubility and solubility product.

had 4 flask A B C & D, and saturated Ca(OH)2 with the following:

Flask A- Ca(OH)2 + 100mL of distilled water

Flask B - Ca(OH)2 + 100mL of 0.0125M NaOH

Flask C- Ca(OH)2 + 100mL of 0.025M NaOH

Flask D -Ca(OH)2 + 100mL of 0.050M Naoh

and titrated 25mL of the solution from each flask with 0.1 M standardized HCL solution.
the amount of HCl hat was required to neutralize each flask (in my case ) was

flaskA- 7 mL
flaskB- 10mL
flaskC- 11mL
flask D- 12mL

The question being asked is to calculate the equilibrium concentrations of OH- and Ca2+ in each case, and then to calculate Ksp.

Please explain how to do the question.

Explanation / Answer

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

Ksp = [Ca2+][OH-]^2

[Ca2+]= 1/2 [OH-]

So, if we know the [OH-] , we can calculate [Ca2+] and the Ksp

Flask A : volume of HCl used = 7ml

Molarity of HCl = 0.1M

No of moles = (0.1mole/ 1000ml)*7ml

= 0.0007mole

HCl + OH- -------> H2O + Cl-

1mole of H+ react with 1mole of OH-

Therefore 0.0007mole of HCl react with 0.0007 mole of OH-

Volume of sample solution taken for rotation = 25 ml

Therefore, concentration of OH- = (0.0007mole/25ml)*1000ml = 0.028M

Therefore, concentration of Ca2+ = 0.028/2 =0.014

Now, we got [Ca2+] and [OH-]

Ksp = [Ca2+][OH-]^2

= 0.014 * (0.028)^2

= 1.1*10^-5

Flask B

Volume of HCl used = 10ml

Mole of HCl used = 0.001

Mole of OH- reacted = 0.001

Now, delete the no of moles OH- due to NaOH

OH- mole due to NaOH =( 0.0125mole/1000)*25= 0.0003125

OH- mole = 0.001 - 0.0003125 = 0.0006875

[OH-] = 0.0275M

[Ca2+] = 0.01375M

Ksp = 0.01375 * 0.0275*0.0275

= 1.04*10^-5

Like wise you can proceed for other flasks

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