In: Chemistry
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.
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