If a slightly soluble electrolyte dissociates according to the equation
Am Kn n K ® m+ + m An -,
the precipitate saturated solution
then the expression for the product of solubility (S) has the form
PR(Kn Am) = a× a = (f+ ×[Km+])n×(f– ×[An -]m).
PR dilute solutions can be expressed using equilibrium concentrations of the ions
PR(Kn Am) = [Km+]n ×[An -]m .
The product of solubility of some salts is given in table. 5 of the application.
If the product of concentrations of ions in solution is greater than the table value of Prali, in solution will be present a precipitate of this substance. Conversely, if the product of concentrations of ions in the solution was less than the table, then the precipitate of the substance is dissolved.
Proli characterizes the solubility of a solid electrolyte at a given temperature: two of the same type of compounds has the greater solubility of the PR, which is more.
The equilibrium molar concentration of ions of Km+ and An - L is proportional to the solubility (mol/dm3) of a substance Kn Am:
[Km+] = L and n [An-] = m L .
Hence, PR(Am, Kn) = (n L)n × ( m-L)m and L = .
Example 1.To determine the PR of magnesium fluoride if its solubility in water of 0.001 mol/dm3.
R e W e n e
MgF2(TV) Û Mg2+ + 2F –м
PR(MgF2) = [Mg2+] [ F –]2 = L ×(2L)2 = 4 L3 = 4 × (0,001)3 = 4 × 10-9.
Example 2.Calculate the equilibrium molar concentration (mol/dm3) anions in a saturated solution of carbonate of silver (I) at 25 °C, if PR = 8,7×10-12.
R e W e n e
Ag2CO3(TV) Û2Ag+ + CO32-;
PR(Ag2CO3) = [Ag+]2 × [CO32- ] = (2 × [CO32- ])2 × [CO32- ] = 4 × [CO32- ]3;
[CO32- ] = 1.3 x 10-4 mol/dm3.
Example 3.To determine the pH of a saturated solution of calcium hydroxide at 25 °C, if PR = 6,3×10-6.
The decision
CA(Oh)2(TV)Û Ca2+ + 2ON– ;
pH = [Ca2+] × [OH–]2 = 1/2×[OH–] × [OH–]2 = 1/2×[OH–]3;
pH = 14 – Ron = 14 + lg[OH–] = 14 + lg = 14 + lg = 12,4.
Example 4.Hitting any precipitate of lead iodide (II) at 25 °C after draining 100 cm3 of a 0.005 M solution of lead nitrate (II) and 200 cm3 of 0.01 M solution of potassium iodide, if PR (PbI2) = 8,7×10-9
The decision
Pb(NO3)2 + length 2KI = PbI2(TV)+ 2KNO3;м
Pb2+ + 2I– = PbI2(TV);
[Pb2+] = ; [I –] = ,
where C1 and C2 – the concentration of ions in the solutions before mixing; V1и V2 – volumes of the initial solutions in the order presented in the problem statement; (V1+ V2) is the volume of the final solution.
By mixing equal volumes of the initial solutions concentration of ions in the final solution is reduced in 2 times in comparison with C1 and C2.
[Pb2+]×[I–]2 == = = 7,4×10-8 .
PbI2осаждается, as the condition of sediment:
Of 7.4×10-8 >is 8.7×10-9.
Z AND D AND CH AND
1. The solubility of AgI equal to 1.2×10-8 mol/dm3. To calculate PR(AgI).
2. 2 dm3 of water at 25 °C soluble of 2.2×10-4 g of silver bromide. To calculate PR(AgBr).
3. Solubility product of PbCl2 is 1.7×10-5. What is the concentration of lead ions in a saturated solution of PbCl2?
4. Solubility product of CaSO4 is equal to 6×10-5. Is there any CaSO4 precipitate by mixing equal volumes of 0.2 n solutions of CaCl2 and Na2CO3?
5. PR (PbI2) = 8,7×10-5. Hitting any residue if you mix equal volumes of solutions containing 3 g/dm3 Pb(NO3)2 and 1 g/dm3 KI?
6. Solubility product of AgCl is 1.6×10-10. Calculate the concentration of a saturated solution of AgCl (in mol/dm3 and g/dm3).
7. How much water is required to dissolve 1 g СаС2О4 at room temperature, if PR(СаС2О4) = 2,6×10-9?
8. How many grams of CaCO3 can be dissolved in 1 dm3 of water at 18 °C, if PR(CaCO3) = 4,8×10-9 at the same temperature?
9. Calculate the concentration of ions Ag+ in a saturated solution of AgBr containing NaBr concentration of 0.01 mol/dm3.
10. To calculate the value of PR, if the solubility of the substance Ме2А in water at a certain temperature equal to 1.2×10-3 mol/dm3.
11. To calculate the value of PR metal hydroxide IU(Oh)2 if pH of its saturated solution is equal to that of 9.54 at 25 °C.
12. To calculate PR(PbSO4), if mass fraction of PbSO4 in a saturated solution at a certain temperature equal to 0,057 % (density of solution is taken equal to 1 g/cm3).
13. To determine whether a precipitate will fall out after draining equal amounts of 0,0023 M solutions of AgNO3 and KBr at 25 °C.
14. To determine whether a precipitate will fall out after draining 5 cm3 of 0.004 M CdCl2 solution and 15 cm3 of 0.003 M solution of NaOH at 25 °C.
15. PR(CA3(PO4)2) = 1×10-25 at 25 °C. Calculate the concentration of Ca2+ and РО43 - in a saturated solution at the same temperature.
16. Saturated at room temperature a solution of PbSO4 volume of 3 dm3 contains 0,132 g of salt. To calculate PR(PbSO4).
17. At 18 °C PR(PbF2) is 3.2×10-8. What is the amount of lead contained in 0.4 dm3 of a saturated solution?
18. A saturated solution AgIO3 volume of 3 dm3 contains in the form of ions 0,176 g of silver. To calculate PR(AgIO3).
19. The solution contains the ions Ba2+ and Sr2+ concentrations respectively equal to 5×10-4 and 5×10-1 mol/dm3. What kind of precipitation will fall in the first solution with the gradual addition of a solution of K2CrO4? PR(SrCrO4) = = 3.6 x 10-5.
20. PR(Ag3PO4) is 1.8×10-18. In what volume of a saturated solution contains 0.05 g of dissolved salt?
21. Hitting any precipitate of calcium sulfate, if to 0.1 dm3 of 0.01 M solution of Ca(NO3)2 added with 0.4 dm3 of 0.001 n H2SO4 solution? The degree of dissociation of Ca(NO3)2 and H2SO4 equal to 95 %; OL(CaSO4) = 6,1×10-5.
22. To calculate PR(Ni(NO3)2), if mass fraction of Ni(NO3)2 in a saturated solution at a certain temperature equal to 0,205 % (density of solution is taken equal to 1 g/cm3).
23. To calculate what volume of water (in dm3) will be required for the dissolution 0,0158 g SrCO3 at 25 °C, if PR(SrCO3) = 5,3×10-10 (the volume of water taken equal to the volume of the solution).
24. Calculate the equilibrium molar concentration (mol/dm3) cations in a saturated solution of salts AgMoO4 (PR = 2,8×10-12) and TlC2O4 (PR = 2,0×10-4) at 25 °C.
25. Calculate the equilibrium molar concentration of anions in a saturated solution of salts BaF2 (S = 1,7×10-6) and Ca(IO3)2 (CR = 1,9×10-6) at 25 °C.