1. calculate the volume of 0.50 M NaOH needed to saturate 100.0 ml of 0.50 M Zn(

Question

1. calculate the volume of 0.50 M NaOH needed to saturate 100.0 ml of 0.50 M Zn(NO3)2 with Zn(OH)2

2. Assuming a 75% recovery of the volume from the first filtration, what volume of cation solution will be used in part B of the procedure? (Hint: you started with 100ml and added NaOH)

3. Calculate the volume of 0.50M NaOH needed to saturate the volume from question 2 of 0.50 M Mn(NO3)2 with Mn(OH)2.

4. Assuming a 75% recovery of the volume from the second fitration, what volume of cation solution will be used in part C of the procedure?

5. Calculate the volume of 0.50M NaOH needed to exceed saturation by 10% the volume from question 4 of 0.50M Mn(NO3)2 with Mn(OH)2.

Explanation / Answer

1.You don't have enough information because I don't want to calculate anything that may use different Ksp than you have in your text/notes.

Look up the Ksp fo Zn(OH)2
(Zn^2+) = Ksp/(OH^-)^2 and solve for Zn^2+ in mols/L, then M x L (100 mL or 0.1L) = mols.

Then Zn^2+ + 2OH^- ==> Zn(OH)2
Convert mols Zn^2+from above to mols NaOH and convert that to mL using M NaOH.

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