Question

Ksp = [Ba²+][10,1 2. 10. mL of 0.10 M Ba(NO), is mixed with 10. mL of 0.10 M KIO,, a precipitate forms. for barium a. Which ion will still be present at appreciable concentration in the equilibrium mixture if K iodate is very small? b. What would that concentration be? 3. Lead chloride, PbCly, is slightly soluble with K equal to 1.7 x 10. a. What is the solubility of lead chloride in pure water? b. What would the solubility be in 0.10 M Nal? c. The difference between these two results is caused by the common ion effect. Stat in words what the common ion effect predicts. 4. When 5.00 ml of 0.120 M Pb(NO), are mixed with 5.00 mL of 0.0300 M KI, a yellow precipitate of Pbl, (s) forms. a. How many moles of Pba are initially present? b. How many moles of l' are initially present? c. The concentration of l' is found to be 8.0 x 10 moles/liter. How many moles of I are present in the solution? d. How many moles of l' precipitated? e. How many moles of Pba precipitated? How many moles of Pb are left in solution? g. What is the concentration of Pba in the equilibrium solution? h. Find a value for K of Pbl, from these data

Answer 1

2. a 10 mL 01 (M) BÀ (NGs) : con no. of miles of Ba - 10 - 10 mol 10m 0.1(M) kiz no. of moles of 10% = 0.1 x 10 103 mol Ba & 10z combines and precipitated from the solution. 1 mol Bat combines with a mol 10z. Here 103 mol Bart and 103 mol tot available. so an excess there will be Bazt, y concentration of Balt: x 103 mol in 20 mL sulention ie 10 x 1000 CM) = 0.025 (M) 3. ksp of PbCl₂ = 1.7x10-5 Pich - P62+ + 20 Let's solerbility=c .: ksp = [p62] [] ex(20) = 903 443 = 1.7x105 c= 0.0162 ay: solubility of Pble in pure water = 0.0162 (M) > O. 1 (M) Nal to ci- (i.e behaves as common ion) It is similars to u (1.8 15 oku.БLty is no c" kep = cх [1-7 = éx (0.1)2 = 1.7x105 c=17X163 () i.e solubility is now decreased.

Yow To Common in effect decreases the solubility of a sparingly soluble salt. 4. 5 mL 0.12 (M) Pb(NO3)2 at no. of moles of P62+ - 0.12 x 5 - 6x10 moles 5 ml of 0.03 M kI: b) no. of moles of I = = 1.5x10 " moles. > Aften mixing [I-] = 8x10 3 moles/litre Total volume of solution=5+5=10 mL . IT present in the solution = 8x103 0.03 *10 1000 = 8x10 5 moles. de precipitated - Buod (1.5X164 - 8x105) miles = 7x105 moles. ex Pb2+ precipitated = 1 of Ir precipitated Cas PbI2) = x 7X105 - 3.5x105 moles. 7 Pb2+ left = 6x109 – 3.7x105 - 5.658 10 4 87 Concentration of P624 - 5.65 x 10 x 1000 (1) 20 = 0.0565 (M)

30 & ksp = [P624] [I -] ? (which is precipitated) = 3.5x105 x 6x175) * +75xto 13 = 2:45 *109