Name _____Mr. Perfect______Date ______F 19______

1. Calculate the pH and percent dissociation for an aqueous 1.5 M HNO2 solution. (10 pts)

-4 Ka = 4.5 x 10 for HNO2 + - HNO2 ⇄ H + NO2 I 1.5 0 0 C -x +x +x E 1.5 – x x x 푥2 푥2 퐾 = 4.5 푥 10−4 = ≈ 푎 1.5 − 푥 1.5 x = 2.60 x 10-2 M = [H+] Check: 2.60 푥 10−2 × 100 = 1.7 % (푝푒푟푐푒푛푡 푑푖푠푠표푐푖푎푡푖표푛) 푎푠푠푢푚푝푡푖표푛 푖푠 푣푎푙푖푑 1.5 pH = -log (2.60 x 10-2) = 1.59

2. Aniline (C6H5NH2) is an organic base used to manufacture dyes. Calculate the pH of a 0.15 M + + solution of aniline. {Hint: C6H5NH2 + H → C6H5NH3 } (10 pts)

-10 Kb = 4.3 x 10 + - C6H5NH2 + H2O ⇄ C6H5NH3 + OH I 0.15 --- 0 0 C -x --- +x +x E 0.15 – x --- x x 푥2 푥2 퐾 = 4.3 푥 10−10 = ≈ 푥 = 8.03 푥 10−6푀 = [푂퐻−] 푏 0.15 − 푥 0.15 pOH = -log(8.03 x 10-6) = 5.10 pH = 14 – 5.10 = 8.90

3. Calculate the pH of a 0.25 M C5H5NHCl salt solution. (10 pts)

-9 Kb =1.7 x 10 for C5H5N + C5H5NH ⇄ C5H5N + H I 0.25 --- 0 0 C -x --- +x +x E 0.25 – x --- x x

−14 2 2 퐾푤 1 푥 10 −6 푥 푥 −3 + 퐾푎 = = −9 = 5.88 푥 10 = ≈ 푥 = 1.21 푥 10 푀 = [퐻 ] 퐾푏 1.7 푥 10 0.25 − 푥 0.25

pH = -log(1.21 x 10-3) = 2.92

Chemistry 102 Exam 2 Name _____Mr. Perfect______Date ______F 19______

4. Carbonic acid (H2CO3) is used to buffer blood. Use the Henderson-Hasselbalch equation to - calculate the ratio of HCO3 to H2CO3 in blood having a pH of 7.40. (10 pts)

-7 -11 Ka1 = 4.3 x 10 and Ka2 = 5.6 x 10 + - H2CO3 ⇄ H + HCO3

[퐴−] 푝퐻 = 푝퐾 + 푙표푔 푎 [퐻퐴] [퐴−] 7.40 = −log (4.3 푥 10−7) + 푙표푔 [퐻퐴]

[퐴−] [퐴−] [퐴−] 7.40 = 6.37 + 푙표푔 1.03 = 푙표푔 표푟 = 101.03 = ퟏퟎ. ퟕ [퐻퐴] [퐻퐴] [퐻퐴]

-6 5. Methyl red is an indicator used in titrations and has a Ka equal to 7.9 x 10 . At what pH will this indicator change colors? Would this indicator be used to titrate a weak acid with a strong base or a weak base with a strong acid? What is the color of this indicator a pH 4.6? (10 pts) HIn ⇄ H+ + In- (Red) (Yellow)

-6 The indicator changes colors at pH = pKa = -log(7.9 x 10 ) = 5.1

This indicator is used for a weak base – strong acid titration. (Equivalence point pH < 7)

At pH 4.6 the color of the solution will appear red.

6. Calculate the solubility (g/L) of MgF2 in solution. (10 pts) -11 Ksp = 7.4 x 10 Mw = 62 g/mol

2+ - MgF2(s) ⇄ Mg (aq) + 2F (aq) I --- 0 0 C --- +s +2s E --- s 2s -11 2 3 Ksp = 7.4 x 10 = s(2s) = 4s s = 2.64 x 10-4 mol/L 푚표푙 62 푔 2.64 푥10−4 × = ퟏ. ퟔퟒ 풙 ퟏퟎ−ퟐ품/푳 퐿 푚표푙

Chemistry 102 Exam 2 Name _____Mr. Perfect______Date ______F 19______

7. A 25.00 mL weak-base sample of 0.15 M aniline (C6H5NH2) is titrated with 0.10 M HCl. Calculate the pH after the following additions of HCl: (15 pts) -10 -3 Kb = 4.3 x 10 aniline: 0.025 L x 0.15 mol/L = 3.75 x 10 mol base a. 10.00 mL of HCl. (before equivalence point) HCl: 0.010 L x 0.10 mol/L = 1.00 x 10-3 mol acid 3.75 x 10-3 mol – 1.00 x 10-3 mol = 2.75 x 10-3 mol base remains

−14 퐾푤 1 푥 10 −5 퐾푎 = = −10 = 2.3 푥 10 퐾푏 4.3 푥 10

(2.75 푥 10−3) 푝퐻 = − log(2.3 푥 10−5) + 푙표푔 = ퟓ. ퟎퟕ (1.00 푥 10−3)

b. 37.5 mL of HCl. (at equivalence point) molesa = molesb 0.0375 L x 0.10 mol/L = 3.75 x 10-3 mol acid 3.75 푥 10−3푚표푙 [퐻퐴] = = 6.00 푥 10−2푀 0.0625 퐿

+ + C6H5NH3 ⇄ C6H5NH2 + H I 0.06 0 0 C -x +x +x E 0.06 – x x x

푥2 푥2 퐾 = 2.3 푥 10−5 = ≈ 푥 = 1.17 푥 10−3 푀 = [퐻+] 푎 0.06 − 푥 0.06 Check: 1.17 푥 10−3 푥100 = 2.0 % 푝퐻 = − log(1.17 푥 10−3 ) = ퟐ. ퟗퟑ 6.00 푥 10−2 c. 45.00 mL of HCl. (past equivalence point)

0.045 L x 0.10 mol /L = 4.50 x 10-3 mol acid

4.50 x 10-3 – 3.75 x 10-3 mol = 7.50 x 10-4 mol HCl remains

7.50 푥 10−4푚표푙 [퐻+] = = 1.07 푥10−2푀 0.070 퐿 pH = -log(1.07 x 10-2) = 1.97

Chemistry 102 Exam 2 Name _____Mr. Perfect______Date ______F 19______

8. Complete the following table. (10 pts)

[H+] [OH-] pH Acid or Base? 3.46 x 10-4 2.88 x 10-11 3.46 acid

1.49 x 10-11 6.7 x 10-4 10.8 base

6.31 x 10-11 1.58 x 10-4 10.2 base

7.2 x 10-9 1.39 x 10-6 8.14 base

9. Which of the following indicators would be appropriate for a titration with an equivalence point pH of 4.7? (5 pts)

Ka pKa range Thymol Blue 7.94 x 10-8 7.1 6.1 – 8.1 Bromocresol Green 1.58 x 10-5 4.8 3.8 – 5.8 Bromocresol Green is appropriate.

10. For the following reactions, predict whether an increase in temperature would be favorable or unfavorable. (10 pts) ∆G = ∆H - T∆S (∆G < 0 is favorable)

a) 2H2O2(aq) → 2H2O(l) + O2(g) +∆S favorable

b) PCl3(l) + Cl2(g) → PCl5(s) -∆S unfavorable

c) CO2(s) → CO2(g) +∆S favorable

d) H2O(g) → H2O(l) -∆S unfavorable

11. Extra Credit. A hydrochloric acid (HCl) solution has a pH of 3.34. Will the addition of solid NaCl salt raise the pH, lower the pH, or have no effect? (5 pts) NaCl → Na+ + Cl- (strong electrolyte)

Cl- has no affinity towards water, therefore there will be no effect on the pH upon addition of NaCl salt.

Chemistry 102 Exam 2