chemistry Problem

Question

Calculate the concentration of each ion in the solutions made by mixing the following pairs of solutions. Assume no precipitate forms yet, and assume all volumes are additive. (a) 100.0 mL each of 0.10 M AgNO3 and 0.10 M KCl; (b) 25 mL of 0.10 M Na2SO4 with 75 mL of 0.20 M BaCl2; (c) 150 mL of 0.010 M Mg(NO3)2 with 250 mL of 0.0010 M LiF; (d) 65 mL of 0.010 M Pb(NO3)2 with 125 mL of 0.025 M NaCl; (e) 75 mL of 0.0010 M CaCl2 with 95 mL of 0.025 M K2SO4; (f) 25.0 mL each of 0.10 M Ca(NO3)2 and 0.10 M Na3PO4. Consult the table of solubility product constants at the end of the chapter and calculate Qsp for any pairs of ions in Exercise 14 that might form a precipitate. Compare Qsp to Ksp and decide if a precipitate will form

Answer 1

(a). 100 ml 0.1 M AgNO₃ + 100 ml 0.1 M KCl

Final concentrations: [Ag⁺] = 0.05 M; [NO₃^-] = 0.05 M; [K⁺] = 0.05 M; [Cl^-] = 0.05 M

(b). 25 ml 0.1 M Na2SO4 + 75 ml 0.2 M BaCl₂

(25 ml )(0.1 M Na₂SO₄) = (100 ml)(x M) and x = 0.025 M Na₂SO₄ = 0.05 M Na⁺ and 0.025 M SO₄^2-

(75 ml)(0.2 M BaCl₂) = (100 ml)(x M) and x = 0.15 M BaCl₂ = 0.15 M Ba²⁺ and 0.30 M Cl^-

(c). 150 ml 0.01 M Mg(NO₃)₂ + 250 ml 0.001 M LiF

(150 ml)(0.01 M Mg(NO₃)₂) = (400 ml)(x M) and x = 0.00375 M Mg(NO₃)₂ = 0.00375 M Mg²⁺ and 0.0075 M NO₃^-

(d). 65 ml 0.01 M Pb(NO₃)₂ + 125 ml 0.025 M NaCl

(65 ml)(0.01 M Pb(NO₃)₂) = (190 ml)(x M) and x = 0.0034 M Pb(NO₃)₂ = 0.0034 M Pb²⁺ and 0.0068 M NO₃^-

(125 ml)(0.025 M NaCl) = (190 ml)(x M) and x = 0.016 M NaCl = 0.016 M Na⁺ and 0.016 M Cl^-

Do the others the same way being careful to double the concentration if there are 2 of a particular ion, triple it for Na in Na₃PO₄.

We don't have the table of Ksp values so cannot address the rest of the question.

Answer 2

To calculate the concentration of each ion in the solutions, we'll use the dilution equation and assume all volumes are additive. The dilution equation is: C1V1 = C2V2 where C1 and V1 are the initial concentration and volume, and C2 and V2 are the final concentration and volume.

(a) 100.0 mL each of 0.10 M AgNO3 and 0.10 M KCl: Since AgNO3 and KCl dissociate completely, the final volume will be 200.0 mL.

For Ag+: C1V1 = C2V2 (0.10 M)(100.0 mL) = C2(200.0 mL) C2 = 0.05 M

For Cl-: C1V1 = C2V2 (0.10 M)(100.0 mL) = C2(200.0 mL) C2 = 0.05 M

(b) 25 mL of 0.10 M Na2SO4 with 75 mL of 0.20 M BaCl2: The final volume will be 100 mL.

For Na+: C1V1 = C2V2 (0.10 M)(25 mL) = C2(100 mL) C2 = 0.025 M

For SO4^2-: C1V1 = C2V2 (0.10 M)(25 mL) = C2(100 mL) C2 = 0.025 M

For Ba^2+: C1V1 = C2V2 (0.20 M)(75 mL) = C2(100 mL) C2 = 0.15 M

For Cl-: C1V1 = C2V2 (0.20 M)(75 mL) = C2(100 mL) C2 = 0.15 M



Follow these steps for parts c,d,e,&f.