The K sp for BaF 2 is 2.4 ´ 10 –5 . When 10 mL of 0.01 M NaF is mixed with 10 mL

Question

           The Ksp for BaF2 is 2.4 ´ 10–5. When 10 mL of 0.01 M NaF is mixed with 10 mL of 0.01 M BaNO3, will a precipitate form? Explan your answer. (2%)

2)       How many moles of Ca(NO3)2 must be added to 1.0 L of a 0.296 M KF solution to begin precipitation of CaF2? For CaF2, Ksp = 4.0 ´ 10–11.(2%)

3)       The Ksp for Mn(OH)2 is 2.0 ´ 10–13. At what pH will Mn(OH)2 begin to precipitate from a solution in which the initial concentration of Mn2+ is 0.10 M? (2%)

4)       Sodium chloride is added slowly to a solution that is 0.010 M in Cu+, Ag+, and Au+. The Ksp values for the chloride salts are 1.9 ´ 10–7, 1.6 ´ 10–10, and 2.0 ´ 10–13, respectively. Which compound will precipitate first? (1%)

5)       The concentration of Mg2+ in seawater is 0.052 M. At what pH will 99% of the Mg2+ be precipitated as the hydroxide? [Ksp for Mg(OH)2 = 8.9 ´ 10–12] (3%)

Explanation / Answer

1) moles F- = 0.010 L x 0.01 M= 0.00010
moles Ba2è = 0.010 L x 0.01 M = 0.00010

total volume = 0.020 L

[F-]= [Ba2+]= 0.00010/ 0.020 L=0.0050 M

Qsp = [Ba2+][F-]^2 = 0.0050( 0.0050)^2=1.2 x 10^-7 < Ksp
precipitation should not occur

2) The precipitation reaction is as follows:
Ca2+(aq) + 2F-(aq) CaF2(s)

Hence the solubility equilibrium is given by
Ksp = [Ca2+] · [F-]²
([...] concentrations in M=mol/L)

The fluoride ions came from dissociation of the KF. Since it is a strong electrolyte you have complete dissociation, where one fluoride ion is formed from dissociation of one molecule of KF: Hence the concentration of fluoride ions is equal to the amount of dissolved potassium fluoride.

[F-] = [KF] = 0.296M

Hence the critical concentration of calcium ions, when percipitation starts, is:
[Ca2+] = Ksp [F-]²
= 4.0×10-11 /(0.296)2
= 4.5×10-10 mol/L

Calcium nitrate contains one calcium ion. Therefore if you add mor than 4.5 × 10-10 moles per liter calciumflouride will precipitate.

3) Ksp=[Mn2+] x [OH-]^2
[OH-]= ( Ksp/ [Mn2+] )0.5
= 1.4142x10^-6

pOH = - log 1.41 x 10^-6 =5.854
pH = 14 - 5.854 = 8.1

4) AuCl. At equal concentrations, the compound with the lowest Ksp value will always precipitate first, which is AuCl in this case

AuCl followed by AgCl and in the last CuCl

5) 99 x 0.052 / 100=0.051 M

8.9 x 10^-12 = 0.051 x [OH-]^2
[OH-] = 1.3 x 10^-5 M
pOH = - log 1.3 x 10^-5 =4.9
pH = 14 - 4.9 = 9.1

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