Please provide a detail answer so that I can better understand how this is done.

Question

Please provide a detail answer so that I can better understand how this is done. Thank you!

a) What is the mass of NaCl that is present in the 'Unknown NaCl' solution?

b) If KCL(aq) was added to the uknown NaCl(aq) solution, would the solubility of Nacl chnage due to the common ion effect? Yes/No and provide explination.

Molarity of AgC(aq)l is 1.330e-5

The Ksp for Silver Chloride is: 1.830e-10.

Buret Reading AgCl Ml V tot NaCl (aq) mL [Ag+] M 0.00 ml 100.0 mL NA 0.50 ml 100.5 ml 1.165e-10 1.50ml 101.5 ml 1.176e-10 3.00ml 103.0 ml 1.194e-10 5.00 ml 105.0ml 1.246e-10 10.50 ml 110.5 ml 1.281e-10 14.00 ml 114.0 ml 1.321e-10 18.00 ml 118.0 ml 1.367e-10 22.50 ml 122.5 ml 1.420e-10 27.50 ml 127.5 ml 1.478e-10 33.00 ml 133.0 ml 1.541e-10 39.00 ml 139.0 ml 1.611e-10 45.00 ml 145.5 ml 1.686e-10

Explanation / Answer

determine the [Ag+] in the solution.
Now consider a saturated solution of AgCl only.
Ksp AgCl = 1.830*10^-10
Ksp = [Ag+][Cl-]
[Ag+] [Cl-] = 1.83*10^-10
[Ag+] = [Cl-] = sqrt(1.83*10^-10)= 1.35*10^-5 M

From our analysis we found that [Ag+] in solution was 1.686E-10M

But [Ag+] [Cl-] = 1.83*10^-10
If [Ag+] = 1.686E-10M , solve for [Cl-]

[Cl-] = (1.83*10^-10) / 1.68610^-10

[Cl-] = 1.085 M That is the total Cl- in solution

You know that the AgCl is contribution 1.686E-10M as Cl- ( because [Ag+] = [Cl-] from AgCl

Therefore the concentration of Cl- from the NaCl =1.085 M - 1.686E-10M) =1.085M

And [NaCl] in solution = 1.085M

so mass of NaCl= 1.085 x .1 x 58.5

=6.349 gm (ANS)

B) IFKCl is added to the solution of NaCl ther will not be any change of solubility of NaCl because KCl is highly soluble in water so it will not precipitate thus there will not be any common ion effect

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