document This contains extra practice problems involving equilibrium, acid/base, buffer, and titration calculations. The exam will also include short answer conceptual questions on these topics.
1. A 0.0560 g quantity of CH3COOH is dissolved in enough water to make 50.0 mL of solution. Calculate the concentrations of H+, CH3COO- and CH3COOH at equilibrium. Ka=1.8 x 10-5.
CH3COOH=60 g/mol 0.0560 g=9.33 x 10-4 mol 9.33 x 10-4 mol/0.050 L=0.01867 M CH3COOH CH3COOH 0.01867 -x 0.01867-x CH3COO0 +x x H3O+ 0 +x x
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Ka=[ CH3COO-][ H3O+]/[CH3COOH] X=5.786 x 10-4 [H3O+]=[ CH3COO-]=5.786 x 10-4 M [CH3COOH]=0.0181 M
2. A titration in conducted where exactly 500 mL of 0.167 M NaOH was titrated into 500 mL 0.100 M CH3COOH. Calculate the pH of the solution at this point in the titration. CH3COOH Ka=1.8 x 10-5. Initial moles NaOH: 0.5 L x 0.167 mol/L=0.0835 mol CH3COOH: 0.5 L x 0.10 mol/L=0.050 mol CH3COOH + -OH CH3COO - + H2O All acid is neutralized, 0.0335 mol -OH remaining 0.0335 mol/L=0.0335 M pOH=1.47 pH=12.53
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3. What is the pH at the equivalence point in the titration discussed in the previous question? 500 mL of 0.100 M CH3COOH =0.05 moles CH3COOH, required 0.05 moles HO0.05 moles HO- x (1 L/0.167 moles)=0.300 L NaOH required Volume at the equivalence point is 0.800 L Concentration of CH3COO - at equivalence point is 0.0625 M Reverse equilibrium: CH3COO - + H2O - CH3COOH + HOICE table, kb=5.55 x 10-10 X=5.89 x 10-6=[HO-] pOH=5.22, pH=8.77
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4. Consider the following equilibrium process at 700C 2 H2 (g) + S2 (g) 2 H2S (g)
Analysis shows that there are 2.50 moles of H2, 1.35 x 10-5 moles of S2, and 8.70 moles of H2S present in a 12.0 L flask at equilibrium. Calculate the equilibrium constant Kc for the reaction.
Concentrations [S2]=1.125 [H2]=0.208 x 10-6 [H2S]=0.725 Kc=[H2S]2 / [H2]2[S2] Kc=1.08 x 107
5. A 2.50 mole quantity of NOCl was initially in a 1.50 L reaction chamber at 400C. After equilibrium was established, it was found that 28 percent of the NOCl had dissociated: 2 NOCl (g) 2 NO (g) + Cl2 (g)
Calculate the equilibrium constant Kc for the reaction.
28% of 2.5 mol of NOCl=0.7 mol reacted, 1.8 mole remaining 0.7 mol NOCl reacted=0.7 mol NO and 0.35 Cl2 mole produced Concentrations 1.8 mole NOCl/1.5 L=1.2 M 0.7 mol NO/1.5 L=0.467 M 0.35 mol Cl2/1.5 L=0.233 M Kc=[Cl2][NO]2 / [NOCl]2 Kc=0.0356
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6. The concentration of HO- ions in a certain household ammonia cleaning solution is 0.0025 M. Calculate the concentration of H+ ions. Ammonia Kb=1.8 x 10-5
[HO-][H+]=1 x 10-14 0.0025 [H+]=1 x 10-14 [H+]=4 x 10-12
7. Calculate the pH of a 0.15 M solution of CH3COONa. What is the percent ionization? Kb=5.6 x 10-10 CH3COO- + H2O CH3COO0.15 -x 0.15-x CH3COOH + HOCH3COOH 0 +x x HO0 +x x
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Kb= [CH3COOH][ HO-] / [CH3COO-] 5.6 x 10-10=x2 / 0.15 X=[ HO-] =9.165 x 10-6 pOH=5.04 pH=8.96 % ionization=9.165 x 10-6 / 0.15=0.0061%
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8. The pH of a 0.30 M solution of a weak base is 10.66. What is the Kb of the base?
B + H2O
BH+ + -OH
pH =10.66, pOH=3.34, [HO]=4.57 x 10-4 B 0.30 -x 0.30-x BH+ 0 +x X HO0 +x 4.57 x 10-4
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This tells us what x is equal to 4.57 x 10-4. Equilibrium concentrations are [B]=0.30 (no significant change), [BH+]=4.57 x 10-4 , [HO-]=4.57 x 10-4 Kb=[BH+][HO-] / [B] = (4.57 x 10-4)2 / 0.30 = 7.1 x 10-7
9. Calculate the pH of a buffer solution of 0.10 M Na2HPO4/0.15 M NaH2PO4. Use the pka of H2PO4- since this is the acid of the buffer, pka=7.21 pH = pKa + log [base]/[acid] pH=7.21 + log (0.10/0.15)=7.03
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