Chemistry 210, March 2012 ANSWERS Extra acid/base/salt equilibrium problems. See solutions beginning on the next page.

Transcription

1 Chemistry 210, March 2012 ANSWERS Extra acid/base/salt equilibrium problems See solutions beginning on the next page. Determine the ph of a M solution of the following substances. CHCl 2 COONa KHS HClO 4 H 2 CO 3 BaCl 2 NH 3 NaF Steps for solving acid/base equilibria: 1. Write a chemical equation of what is or could be happening * For ionic compounds, consider each ion separately first. * For ionic compounds, leave spectator ions out of the equation. 2. If needed (and it probably is needed), fill in an ICE table (initial, change, equilibrium) with given and deduced information. 3. Write the appropriate equilibrium constant expression, or. Make sure to get the correct value of or if needed. 4. Determine what needs to be solved for and solve making simplifying assumptions as appropriate. 5. Check your answer to see if it makes sense. Solutions to practice problems begin on the next page.

2 Solutions: a. Determine the ph of a M solution of CHCl 2 COONa The Na + ion will not react with water. The CHCl 2 COO ion is the conjugate base of a weak acid (similar to acetic acid) so it will be basic and will react with water to form the conjugate weak acid and OH. CHCl 2 COO (aq) + H 2 O(l) CHCl 2 COOH(aq) + OH (aq) The anion is acting as a base so we need to find the for this base. The for the conjugate acid is found in ChemActivity 46 I think (or online where I found it) as 5.5 x K w 1.00x10"14 5.5x10 "2 1.8x10 "13 [ [ CHCl 2 COO [ OH 1.8x10 "13 CHCl 2 COOH x 1.8x10"13 Solve the above system for x. However, since the dichloroacetate ion is a fairly weak base, as evidenced by the very small value, x is probably small compared to M and may be able to be ignored. So, let s check by ignoring x on the bottom of the expression. Yes! 1.9 x rounds off to poh log[oh log(1.9 x 10 7 ) 6.72 ph poh x10"13 ()(1.8x10 "13 ) 3.6x10 "14 x 3.6x10 "14 1.9x10 "7 M [OH Finally, does the answer make sense? Yes, a ph of 7.28 is only slightly basic as would be expected for a very weak base such as the one in this problem.

3 b. Determine the ph of a M solution of KHS. The K + ion will not react with water. The HS ion is the conjugate base of a weak acid, H 2 S, so it will be basic and will react with water to form the conjugate weak acid and OH. HS (aq) + H 2 O(l) H 2 S(aq) + OH (aq) The anion is acting as a base so we need to find the for this base. The for the conjugate acid is found in ChemActivity 46 as 1.0 x K w 1.00x10"14 1.0x10 "7 1.0x10 "7 1.0x10 "7 H 2 S [ [ HS [ OH x 1.0x10"7 Solve the above system for x. However, since the hydrogen sulfide ion is a fairly weak base, as evidenced by the small value, x is probably small compared to M and may be able to be ignored. So, let s check by ignoring x on the bottom of the expression. 1.0x10"7 ()(1.0x10 "7 ) 2.0x10 "8 x 2.0x10 "8 1.4x10 "4 M [OH Yes! 1.4 x rounds off to poh log[oh log(1.4 x 10 4 ) 3.85 ph poh Finally, does the answer make sense? Yes, a ph of is basic as would be expected for a weak base such as the one in this problem.

4 c. Determine the ph of a M solution of HClO 4. Note: This acid is a STRONG acid so an equilibrium calculation is not appropriate. It will essentially fully dissociate producing an [H 3 O + of M. The ph can be calculated directly from this value of [H 3 O +. ph log[h 3 O + log() 0.70 Does the answer make sense? Yes, this is a very acidic solution as would be expected for a M solution of a strong acid. d. Determine the ph of a M solution of H 2 CO 3. H 2 CO 3 is a weak acid. We simply need to write the dissociation for this acid, look up its, and solve for the H 3 O + concentration. H 2 CO 3 (aq) + H 2 O(l) HCO 3 (aq + H 3 O + (aq) The for carbonic acid, H 2 CO 3, I found online as 4.4 x Solve the above system for x. However, since the carbonic acid is a fairly weak acid, as evidenced by the small value, x is probably small compared to M and may be able to be ignored. So, let s check by ignoring x on the bottom of the expression. 4.4x10"7 ()(4.4x10 "7 ) 8.8x10 "8 [ [ H 3 O + 4.4x10 "7 HCO 3 H 2 CO 3 [ x 8.8x10 "8 3.0x10 "4 M [H 3 O + x 4.4x10"7 Yes! 3.0 x rounds off to ph log[h 3 O + log(3.0 x 10 4 ) 3.53 Finally, does the answer make sense? Yes, a ph of 3.53 is acidic as would be expected for a weak acid such as carbonic acid.

5 e. Determine the ph of a M solution of BaCl 2. Ba 2+ is a group 2 cation and these do not react appreciably with water so it will not change the ph of the solution. Cl is the conjugate base of the STRONG acid, HCl. As such, the base Cl is essentially nonbasic ( is less than ). Since neither ion reacts with water a solution of this salt will have a ph of 7. f. Determine the ph of a M solution of NH 3. NH 3 is a weak base. We simply need to write the dissociation for this base, look up its value, and solve for the OH concentration. The last step is either solving for the poh and then the ph or solving for the [H 3 O + and then the ph. NH 3 (aq) + H 2 O(l) NH + 4 (aq) + OH (aq) The for NH 3 is found in the workbook or other source is 1.8 x x10 "5 NH + 4 [ [ OH [ NH 3 Solve the above system for x. However, since ammonia is a weak base, as evidenced by the small value, x is probably small compared to M and may be able to be ignored. So, let s check by ignoring x on the bottom of the expression. Yes! 1.9 x rounds off to 1.8x10"5 ()(1.8x10 "5 ) 3.6x10 "6 x 1.8x10"5 x 3.6x10 "6 1.9x10 "3 M [OH Note that out of is about 1% but 1% of a number with only two sig figs like is not enough to change it. poh log[oh log(1.9 x 10 3 ) 2.72 ph poh Finally, does the answer make sense? Yes, a ph of is basic as would be expected for a weak base such as ammonia.

6 g. Determine the ph of a M solution of NaF. The Na + ion will not react with water. The F ion is the conjugate base of a weak acid, HF, so it will be basic and will react with water to form the conjugate weak acid and OH. F (aq) + H 2 O(l) HF(aq) + OH (aq) The anion is acting as a base so we need to find the for this base. The for the conjugate acid is found in your workbook, textbook or online as 6.6 x K w 1.00x10"14 6.6x10 "4 1.5x10 "11 1.5x10 "11 HF [ [ F [ OH Solve the above system for x. However, since the fluoride ion is a fairly weak base, as evidenced by the small value, x is probably small compared to M and may be able to be ignored. So, let s check by ignoring x on the bottom of the expression. Yes! 1.7 x rounds off to 1.5x10"11 ()(1.5x10 "11 ) 3.0x10 "12 x 1.5x10"11 x 3.0x10 "12 1.7x10 "6 M [OH poh log[oh log(1.7 x 10 6 ) 5.76 ph poh Finally, does the answer make sense? Yes, a ph of 8.24 is slightly basic as would be expected for a weak base such as the fluoride ion.