AP Chem Unit 8

A student mixes 40.mL of 0.10MHBr(aq) with 60.mL of 0.10MKOH(aq) at 25°C. What is the [OH−] of the resulting solution? A [OH−]=0.060M B [OH−]=0.033M C [OH−]=0.020M D [OH−]=0.00000010M

C.

A student prepares a lactic acid-sodium lactate buffer solution by mixing 40.mL of 0.50MHC3H5O3(aq) with 200.mL of 1.0MNaC3H5O3(aq). The pKa of HC3H5O3 is 3.08. What is the pHof the resulting solution? A 2.08 B 3.08 C 3.38 D 4.08

D 4.08

Which of the following is the correct mathematical expression to use to calculate the pH of a 0.10M aqueous Ba(OH)2 solution at 25°C? A pH=−log(0.10) B pH=−log(0.20) C pH=14.00−log(0.10) D pH=14.00+log(0.20)

D pH=14.00+log(0.20)

CH3COOH(aq)+H2O(l)⇄H3O+(aq)+CH3COO−(aq) CH3COO−(aq)+H2O(l)⇄CH3COOH(aq)+OH−(aq) 2H2O(l)⇄H3O+(aq)+OH−(aq) A buffer solution is made up of acetic acid (CH3COOH) and sodium acetate (NaCH3COO). The major equilibria in the buffer system are represented above. Which of the following equilibria could be used to support the claim that the addition of a small amount of NaOH to the buffer will result in only a very small change in pH? A CH3COOH(aq)+CH3COO−(aq)⇄CH3COO−(aq)+CH3COOH(aq) B CH3COOH(aq)+OH−(aq)⇄CH3COO−(aq)+H2O(l) C H3O+(aq)+OH−(aq)⇄2H2O(l) D CH3COOH(aq)+H2O(l)⇄H3O+(aq)+CH3COO−(aq)

B CH3COOH(aq)+OH−(aq)⇄CH3COO−(aq)+H2O(l)

A student mixes 40.mL of 0.10MHBr(aq) with 60.mL of 0.10MKOH(aq) at 25°C. What is the [OH−] of the resulting solution? A [OH−]=0.060M B [OH−]=0.033M C [OH−]=0.020M D [OH−]=0.00000010M

C [OH−]=0.020M

B(aq)+H2O(l)⇄HB+(aq)+OH−(aq)Kb=[HB+][OH−]/[B] The table above provides the chemical structures for weak bases and their ionization constants, Kb. Based on the data, which of the following provides the best reason for the trend in base strengths? A The number of hydrogen atoms B The number of resonance structures C The different electronegativities of H, I, and Br D The different molar masses

c. The different electronegativities of H, I, and Br

HNO2(aq)+NH3(aq)⇄NH4+(aq)+NO2−(aq)Kc=1×106 Nitrous acid reacts with ammonia according to the balanced chemical equation shown above. If 50.mL of 0.20MHNO2(aq) and 50.mL of 0.20MNH3(aq) are mixed and allowed to reach equilibrium at 25°C, what is the approximate [NH3] at equilibrium? A 0.00010M B 0.0010M C 0.010M D 0.10M

A.

HCN(aq)+H2O(l)⇄H3O+(aq)+CN−(aq) Ka=[H3O+][CN−]/[HCN]=6.2×10^−10 The equilibrium reaction shown above represents the partial ionization of the weak acid HCN(aq). A 0.200MHCN(aq) solution has a pH≈4.95. If 0.05g (0.010mol) of NaCN(s) is added to 100mL of 0.200MHCN(aq), which of the following explains how and why the pH of the solution changes? A The pH will be higher than 4.95 because adding CN− will disrupt the equilibrium, resulting in an increased production of HCN that decreases the concentration of H3O+. B The pH will be lower than 4.95 because adding CN− will disrupt the equilibrium, resulting in an increased production of HCN that decreases the concentration of H3O+. C The pH will be higher than 4.95 because CN− is a strong base that can neutralize HCN. D The pH will remain close to 4.95 because the Ka is so small that hardly any products form.

A The pH will be higher than 4.95 because adding CN− will disrupt the equilibrium, resulting in an increased production of HCN that decreases the concentration of H3O+.

A 40.0mL sample of the weak base C5H11N was titrated with 1.00MHCl at 25°C. Based on the resulting titration curve shown above, which of the following pairs provide the best estimates for the pKb and Kb of C5H11N? A pKb≈3.0 and Kb≈1×10−3 B pKb≈5.8 and Kb≈2×10−6 C pKb≈11.0 and Kb≈1×10−11 D pKb≈12.5 and Kb≈3×10−2

A pKb≈3.0 and Kb≈1×10−3

Each particle diagram shown is a representation of an aqueous solution of one of the acids listed in the table. The molarity of the acids in the solutions is the same. Based on the information, which particle diagram best corresponds to the indicated acid? A Diagram 1 corresponds to HClO. B Diagram 2 corresponds to HClO. C Diagram 2 corresponds to HIO. D Diagram 3 corresponds to HBrO.

C Diagram 2 corresponds to HIO.

H2O(l)+H2O(l)⇄H3O+(aq)+OH−(aq) At 5.0°C, the value of Kw for the equilibrium shown above is 1.9×10−15 and the value of pKw is 14.73. Based on this information, which of the following is correct for pure water at this temperature? A [H3O+] =Sqrt(1.9×10^−15) B pH=−log(1.9×10^−15) C 14.73=[H3O+]eq[OH−]eq D pOH=pH+14.73

A. [H3O+] =Sqrt(1.9×10^−15)

In pure water, some of the molecules ionize according to the equation H2O→H++OH−. The extent of the ionization increases with temperature. The graph above shows the pH of pure water at different temperatures. Which of the following represents the variations in the pOH of pure water under the same conditions? A B C D

A

B(aq)+HCl(aq)⇄HB+(aq)+Cl−(aq) The reaction of a weak base, B, with HCl is represented by the equation above. The graph shows the titration curve for 25.0mL of an aqueous solution of B titrated with 0.100MHCl. Based on the graph, which of the following best estimates the initial concentration of the solution of the weak base? A 0.20M B 0.10M C 0.067M D 0.048M

A 0.20M

HF(aq)+H2O(l)⇄H3O+(aq)+F−(aq)pKa=3.20at25°C The acid ionization equilibrium for the weak acid HF is represented by the equation above. To prepare a buffer with a pH=3.50, a student needs to mix 250.mL of 0.100MHF and 250.mL of 0.100MKF. If the student mistakenly mixes 250.mL of 0.0500MHF and 250.mL of 0.0500MKF, which of the following is the result of this error? A The buffer will have a lower capacity because of the smaller number of moles of HF and F− available to react if an acid or base is added. B The buffer will have a lower capacity because the smaller amount of HF and F− will lower the pH of the buffer, and buffers of lower pH have a lower buffer capacity. C The buffer will have a higher capacity because a larger proportion of HF and F− will ionize at lower concentrations, resulting in the neutralization of any added acid or base. D The buffer will have the same capacity because the large volume of the buffer solution dilutes any added acid or base.

A The buffer will have a lower capacity because of the smaller number of moles of HF and F− available to react if an acid or base is added.

A student pours a 10.0mL sample of a solution containing HC2H3O2(pKa=4.8) and NaC2H3O2 into a test tube. The student adds a few drops of bromocresol green to the test tube and observes a yellow color, which indicates that the pH of the solution is less than 3.8. Based on this result, which of the following is true about the relative concentrations of HC2H3O2 and NaC2H3O2 in the original solution? A [HC2H3O2]>[NaC2H3O2] B [HC2H3O2]=[NaC2H3O2] C [HC2H3O2]<[NaC2H3O2] D The relative concentrations cannot be determined without knowing the value of pKb for C2H3O2−.

A [HC2H3O2]>[NaC2H3O2]

Which of the following statements about the pH of 0.010MHClO4 is correct? A pH=2.00, because [H+]=1.0×10−2M. B pH=2.00, because [H+]=2.0×10−2M. C pH>2.00, because HClO4 is a strong acid. D pH<2.00, because HClO4 is a weak acid.

A pH=2.00, because [H+]=1.0×10^−2M.

In pure water, some of the molecules ionize according to the equation H2O→H++OH−. The extent of the ionization increases with temperature. A student heats pure water and records the measured pH at 50°C as 6.6. Based on this information, which of the following mathematical relationships gives the pOH of pure water at 50°C? A pOH=pH B pOH=1pH C pOH=14−pH D pOH=1×10−14pH

A. pOH=pH

B(aq)+HCl(aq)⇄HB+(aq)+Cl−(aq) The graph above shows the titration curve for an aqueous solution of a weak base, B, with HCl as the titrant. Based on the graph, which of the following best estimates the pKa of HB+? A 12.0 B 10.8 C 6.0 D 1.8

B 10.8

HC4H7O2(aq)+H2O(l)⇄H3O+(aq)+C4H7O2−(aq) The chemical equation above represents the acid ionization equilibrium for HC4H7O2 for which pKa=4.8. Which of the following is the best estimate for the pH of a buffer prepared by mixing 100.mL of 0.20MHC4H7O2 with 100.mL of 0.10MNaC4H7O2 ? A 1.0 B 4.5 C 4.8 D 7.0

B 4.5

The equilibrium for the reaction between (CH3)2NH, a weak base, and water is represented by the equation below. The table shows the pH of three solutions of (CH3)2NH(aq) at 25°C. (CH3)2NH(aq)+H2O(l)⇄(CH3)2NH2+(aq)+OH−(aq)Kb=5.4×10−4at25°C [(CH3)2NH]pH at25°C0.050 11.69 0.10 11.85 0.20 12.01 A student mixes 100.mL of 0.200M(CH3)2NH(aq) with 100.mL of 0.200M(CH3)2NH2Cl(aq) and claims that if a small amount of strong base is added to the mixture, then the resulting change in pH of the mixture will be smaller than the change in pH that would result from adding the same amount of strong base to 200.mL of 0.200M(CH3)2NH(aq). Which of the following best explains whether or not the student's claim is correct?

B The claim is correct because the mixture contains a significant amount of the acid (CH3)2NH2+(aq), which can react with and partially neutralize the added strong base, thereby reducing the change in pH. In 0.200M(CH3)2NH(aq), the concentration of the acid (CH3)2NH2+(aq) is very small; therefore, the added strong base is not neutralized and the change in pH will be larger.

HF(aq)+H2O(l)⇄H3O+(aq)+F−(aq) The equation above represents the acid ionization equilibrium for HF. To prepare a buffer with pH≈3.50, 4.20g of NaF(s) should be added to 500.0mL of 0.100MHF(aq). The buffer is accidentally prepared using 90% pure NaF(s) instead of 99% pure NaF(s). Assume that the impurities in the NaF(s) samples are inert. Which of the following explains how the error affects the pH and capacity of the buffer? A The pH is slightly higher than 3.50 and it has a lower capacity for the addition of acids because less than 4.20g of NaF(s) was added. B The pH is slightly higher than 3.50 and it has a higher capacity for the addition of acids because more than 4.20g of NaF(s) was added. C The pH is slightly lower than 3.50 and it has a lower capacity for the addition of acids because less than 4.20g of NaF(s) was added. D The pH is slightly lower than 3.50 and it has a higher capacity for the addition of acids because more than 4.20g of NaF(s) was added.

C The pH is slightly lower than 3.50 and it has a lower capacity for the addition of acids because less than 4.20g of NaF(s) was added.

NH3 is a weak base that reacts with water according to the chemical equilibrium represented above. The table provides some information for two NH3(aq) solutions of different concentration at 25°C. Which of the following is true about the more concentrated 0.30MNH3(aq), and why? A [OH−]=3.2×10−3M and pOH<2.78, because a higher [OH−] corresponds to a lower pOH for 0.30MNH3(aq)compared to 0.15MNH3(aq). B [OH−]=3.2×10−3M and pOH>2.78, because a higher [OH−] corresponds to a higher pOH for 0.30MNH3(aq)compared to 0.15MNH3(aq). C [OH−]=2.3×10−3M and pOH<2.78, because a higher [OH−] corresponds to a lower pOH for 0.30MNH3(aq)compared to 0.15MNH3(aq). D [OH−]=2.3×10−3M and pOH>2.78, because a higher [OH−] corresponds to a higher pOH for 0.30MNH3(aq)compared to 0.15MNH3(aq).

C [OH−]=2.3×10−3M and pOH<2.78, because a higher [OH−] corresponds to a lower pOH for 0.30MNH3(aq)compared to 0.15MNH3(aq).

Which of the following pairs of mathematical expressions can be used to correctly calculate the pH and pOH of a 0.0015MKOH(aq) solution at 25°C? A pH=−log(14.00−0.0015) and pOH=−log(0.0015) B pH=log(0.0015)−14.00 and pOH=−log(0.0015) C pH=14.00−(−log(0.0015)) and pOH=−log(0.0015) D pH=−(−log(0.0015)) and pOH=14.00−(−log(0.0015))

C pH=14.00−(−log(0.0015)) and pOH=−log(0.0015)

HF(aq)+H2O(l)⇄H3O+(aq)+F−(aq) Ka=6.3×10−4 at 25°C The acid ionization equilibrium for HF is represented by the chemical equation above. A student claims that the pH of a solution that contains 0.100MHF(aq) and 0.100MNaF(aq) will change only slightly when small amounts of acids or bases are added. Which of the following pairs of equations can the student use to justify the claim? A HF(aq)+OH−(aq)→H2O(l)+F−(aq) and OH−(aq)+H3O+(aq)→2 H2O(l) B H3O+(aq)+OH−(aq)→2 H2O(l) and F−(aq)+H3O+(aq)→HF(aq)+H2O(l) C HF(aq)+OH−(aq)→H2O(l)+F−(aq) and F−(aq)+H2O(l)→HF(aq)+OH−(aq) D HF(aq)+OH−(aq)→H2O(l)+F−(aq) and F−(aq)+H3O+(aq)→HF(aq)+H2O(l)

D HF(aq)+OH−(aq)→H2O(l)+F−(aq) and F−(aq)+H3O+(aq)→HF(aq)+H2O(l)

The equilibrium for the reaction between (CH3)2NH, a weak base, and water is represented by the equation below. The table shows the pH of three solutions of (CH3)2NH(aq) at 25°C. (CH3)2NH(aq)+H2O(l)⇄(CH3)2NH2+(aq)+OH−(aq)Kb=5.4×10−4at25°C [(CH3)2NH]pH at25°C0.050 11.69 0.10 11.85 0.20 12.01 Based on the information given, which of the following is true? A Solutions with a higher concentration of (CH3)2NH have a higher pOH because to reach equilibrium a smaller amount of the conjugate acid (CH3)2NH2+ is produced. B Solutions with a higher concentration of (CH3)2NH have a higher pOH because to reach equilibrium more OH− is produced. C Solutions with a higher concentration of (CH3)2NH have a higher pH because to reach equilibrium a smaller amount of the conjugate acid (CH3)2NH2+ is produced. D Solutions with a higher concentration of (CH3)2NH have a higher pH because to reach equilibrium more OH− is produced.

D Solutions with a higher concentration of (CH3)2NH have a higher pH because to reach equilibrium more OH− is produced.

HC2H3O2(aq)+H2O(l)⇄H3O+(aq)+C2H3O2−(aq) pKa=4.76 The equilibrium for the acid ionization of HC2H3O2 is represented by the equation above. A student wants to prepare a buffer with a pH of 4.76 by combining 25.00mL of 0.30MHC2H3O2 with 75.00mL of 0.10MNaC2H3O2. While preparing the buffer, the student incorrectly measures the volume of NaC2H3O2 so that the actual volume used is 76.00mL instead of 75.00mL. Based on the error, which of the following is true about the buffer prepared by the student? A The pH of the buffer will be slightly lower than 4.76because the total volume of the buffer is 101.00mLinstead of 100.00mL, and the HC2H3O2 was diluted. B The pH of the buffer will be slightly lower than 4.76because the amount of C2H3O2− added was higher than the amount of HC2H3O2 added. C The buffer solution will have a slightly higher capacity for the addition of bases than for the addition of acids because the total volume of the buffer is 101.00mLinstead of 100.00mL, and the HC2H3O2 was diluted. D The buffer solution will have a slightly higher capacity for the addition of acids than for the addition of bases because the amount of C2H3O2− added was higher than the amount of HC2H3O2 added.

D The buffer solution will have a slightly higher capacity for the addition of acids than for the addition of bases because the amount of C2H3O2− added was higher than the amount of HC2H3O2 added.