1)A solution is prepared by mixing 50.0 mL of 0.33 M Pb(NO 3 ) 2 with 50.0 mL of
ID: 484115 • Letter: 1
Question
1)A solution is prepared by mixing 50.0 mL of 0.33 M Pb(NO3)2 with 50.0 mL of 1.0 M KCl. Calculate the concentration of Cl- at equilibrium, Ksp for PbCl2(s) is 1.6 × 10-5.
2)Calculate the NH4+ concentration that is needed to prevent Mg(OH)2 from precipitating from a liter of solution which contains (1.000x10^-2) M ammonia and (2.9x10^-3) M of Mg2+ (aq). (Ka (NH4+) = 5.6 x 10–10, Ksp (Mg(OH)2) = 1.12 x 10–11)
3)A solution is saturated with SrCO3 (Ksp = 7.0 x 10–10) and SrF2 (Ksp = 7.9 x 10–10). The carbonate concentration is found to be (1.7x10^-3) M. What is the fluoride concentration?
Explanation / Answer
1)A solution is prepared by mixing 50.0 mL of 0.33 M Pb(NO3)2 with 50.0 mL of 1.0 M KCl. Calculate the concentration of Cl- at equilibrium, Ksp for PbCl2(s) is 1.6 × 10-5.
Number of moles = molarity * volume in L
50 ml = 0.050 L (because 1000 ml = 1.0L)
The number of molesof Pb2+ ions = 0.33 M * 0.05L
=0.0165
The number of molesof Cl- ions = 1.0 M * 0.05 L
= 0.05
Total volume = 50+50 = 100 mL = 0.1 L
[Pb2+ ] = (0.0165 / 0.1L) =0.165 M
[Cl- ] = (0.05 / 0.1L) = 0.5 M
[Pb2+] = 0.0050 / 0.100 =0.05 M
[Cl-] = 0.080 / 0.100 = 0.80 M
Qsp = ( 0.165)(0.80)^2 =0.1056>> Ksp so precipitation wolud occur
1.6 x 10^-5 = (x) (2x)^2 = 4x^3
x^3 =1.6 x 10^-5/4
x^3 = 4*10^-6
x = 0.0158
x = [Pb2+] = 0.0158 M
[Cl-] = 0.0158 * 2 = 0.0317 M
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