Lecture 32 Acids and Bases IV Tutorial

www.apchemsolutions.com Lecture 32 Acids and Bases IV Tutorial

1) When a 31.0g sample of NaCl is added to a 1.0 L saturated solution of PbCl2 , does the concentration of Pb2+ increase, decrease, or stay the same? Justify your answer. Assume that the overall volume of the solution does not change.

2+ - PbCl2(s) U Pb (aq) + 2 Cl (aq) + - NaCl(s) Æ Na (aq) + Cl (aq) - NaCl(s) will dissolve completely. The addition of Cl (aq) ions will shift the equilibrium of - the first reaction to the left, according to Le Chatelier’s principle. Excess Cl (aq) will 2+ combine with Pb (aq) to form the precipitate PbCl2(s). This will continue until the product 2+ - 2 2+ of [Pb ][Cl ] equals Ksp once again. This will decreases the concentration of Pb in the solution.

2) A buffered solution is created by dissolving 0.30 mol of sodium cyanide, NaCN, -10 in 510 mL of 0.55 M HCN. (Ka = 6.2 x 10 ) Assume that the volume of the solution does not change. a. Find the pH of the solution.

0.30 mol [CN]− (0.59) =+0.59MK CN−− , pH = p log=− log(6.2× 1010 )+ log= 9.2 0.510 L a [HCN] (0.55)

b. Find the pH after 0.25 mol HNO3 is added to the solution. Assume that the volume does not change.

0.55 mol HCN 0.510 L× =0.28 mol HCN 1L H+− + CN → HCN I 0.25 mol 0.30 mol 0.28 mol C -0.25 mol -0.25 mol +0.25 mol E 0 mol 0.05 mol 0.53 mol 0.05 mol 0.53 mol Molarity ==0.1M CN− , Molarity ==1.0M HCN CN- 0.510 L HCN 0.510 L [CN− ] (0.1M ) pH = pK +=−×+= log log(6.2 10−10 ) log 8.2 a [HCN] (1.0M )

© 2009, 2008 AP Chem Solutions. All rights reserved. 1 www.apchemsolutions.com 3) A solution is created by dissolving a 1.786 g sample of potassium acetate, o KCH3CO2, in 750.0 mL of 0.345 M acetic acid, CH3COOH, at 25 C. Ka for acetic acid is 1.8 x 10-5 at 25oC. Assume that the volume of the solution does not change. - a. Find the concentration of the acetate ion, CH3COO , in the final solution.

1 mol CH COO- 1.786g KCH COO×=3 1.820× 10−2 mol CH COO− 3 98.15g KCH COO 3 3 -2 - 1.820×10 mol CH3 COO −2 − Molarity - ==2.4× 10M CH3 COO CH3 COO 0.75 L

b. Find the concentration of acetic acid, CH3COOH, in the final solution.

The final concentration of acetic acid is 0.345 M CH3COOH. We can assume that none - of the acetic acid dissociated, as adding CH3COO will drive the equilibrium to the left to produce more CH3COOH. Furthermore, acetic acid is a weak acid, so, even without the salt, it experiences very little dissociation.

c. Find the concentration of H+ in the final solution

−+ CH33 COOH → CH COO + H IM 0.345 0.024 MM 0 ~ C -x + x + x ~ EM 0.345 - x 0.024 M + x x ~

−+ [CH3 COO ][H ] (0.024 MM + x)(x) (0.024 )(x) Ka == ≈ [CH3 COOH] 0.345 M - x 0.345 M K (0.345) (1.8× 10−5 )(0.345) x2==a =×.610−4 M 0.024 0.024 [H+− ]== x 2.6 × 10 4 M

d. Find the pH of the final solution.

pH = -log[H+− ] = -log(2.6× 104 ) = 3.6

© 2009, 2008 AP Chem Solutions. All rights reserved. 2 www.apchemsolutions.com 4) How many moles of KOCl must be added to a 550 mL solution of 0.35 M HOCl -8 to create a buffed solution with a pH of 8.0? (Ka = 3.5 x 10 ) Assume that the volume does not change.

[H+−− ]===× 10-pH 10 8 1 10 8 M

[H+− ][OCl ] K = a [HOCl] K [HOCl] (3.5× 10−8 )(0.35) [OCl− ] ==a =1.2M HOCl [H+− ] 1× 10 8

1.2 M HOCl n =× 0.55 L =0.66 mol KOCl KOCl 1 L

5) Solid strontium carbonate is placed in a solution of sulfuric acid.

+ 2+ SrCO3(s) + 2 H (aq) Æ Sr (aq) + H2O(l) + CO2(g)

6) Solid beryllium hydroxide is placed in a solution of acetic acid.

2+ - Be(OH)2(s) + 2 CH3COOH(aq) Æ Be (aq) + 2 H2O(l) + 2 CH3COO (aq)

7) Solid silver chloride is sprinkled into a concentrated ammonia solution.

+ - AgCl(s) + 2 NH3 Æ Ag(NH3)2 (aq) + Cl (aq)