Practice Sheet 15 Answers PDF

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Practice Sheet 15 Answers...

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Practice Sheet 15 and Answers Please use these answers (at the end) to make sure you understand and know how to do the problems and to study for the exam. Practice Problem

Parallel Example

15D.1 (a) When 6.3 g of nitric acid are dissolved in water to produce 500.0 mL of solution, what is the resulting pH? (b) When 9.8 g of sulfuric acid are dissolved in water to produce 500.0 mL of solution, in what range will be the resulting pH?

(a) When 3.65 g of hydrochloric acid are dissolved in water to produce 500.0 mL of solution, what is the resulting pH? (b) When 0.98 g of sulfuric acid are dissolved in water to produce 1000.0 mL of solution, in what range will be the resulting pH?

15D.2. An aqueous solution of HNO3 has a pH of 2.34. What is the concentration of the acid?

What is the pH of a 0.040M solution of HClO4?

15D.3. What is the concentration of a solution of (a) KOH for which the pH is 11.89; (b) Ca(OH)2 for which the pH is 11.68?

What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH)2?

15D.4. Your lab partner claims it is possible to make a basic solution by adding an acid to water. She says all you need to do is make a 2.0×10−9 M aqueous solution of HI, giving [H+] = 2.0×10–9 M. Taking the negative log of this concentration yields a pH of 8.7. This is within the range of bases on the pH scale. What is going on here? Can your lab partner really defy the laws of chemistry?

What is the pH of a 1.00×10–8 M solution of HCl? Does your answer make sense? Why or why not? Can you think of a solution to this problem?

Key to Parallel Example (a) 3.65 g of HCl is 0.100 mol HCl. The concentration of HCl is therefore 0.100 mol/0.500 L = 0.2 M. Since HCl is a strong monoprotic acid, if is completely ionized the concentration of hydronium ion will therefore be 0.200 M. The pH is the –log of this value, namely 0.70. (b) 0.98 g of H2SO4 is 0.010 mol H2SO4. The concentration of H2SO4 is therefore 0.010 M. Since H2SO4 is a diprotic acid, we have to consider the concentration of hydronium ion as the sum of hydronium produced from the first ionization step + hydronium produced from the second ionization step. The first ionization is strong (i.e., complete). Therefore, the contribution from the first ionization step is 0.01 M. But the second ionization is weak, so it will contribute less than 0.01 M. So the resulting concentration of hydronium will be between 0.01 and 0.02 M. The pH therefore falls between 2 and 1.70. Because HClO4 is a strong acid, it is completely ionized, giving [H+] = [ClO4-] = 0.040 M. The pH of the solution is given by pH = -log(0.040) = 1.40. (a) NaOH dissociates in water to give one OH- ion per formula unit. Therefore, the OH- concentration equals the stated concentration of NaOH, namely 0.028 M. Thus the pOH is –log(0.028) = 1.55. Because pH + pOH = 14, then pH= 12.45. (b) Ca(OH)2 dissociates in water to give two OH- ions per formula unit. Therefore, the OH- concentration equals twice the stated concentration, namely 0.0022 M. Thus the pOH is –log(0.0022) = 2.96. Because pH + pOH = 14, then pH= 11.04. To find the pH of the solution we need the equation pH = –log[H+]. When we take the −log of the [H+], we get a pH of 8! This does not make sense. A strong acid like HCl should have an acidic pH (less than 7). The reason for our miscalculation is due to the autoprotolysis of water. 2H2O = H3O+ + OH– We know that in pure water [H3O+] = [OH–] = 1×10–7 M, which is larger than our 1×10–8 M HCl solution. To correctly solve this problem, we would need to take into account the ionization of water simultaneously with the dissociation of HCl, which would give a pH of 6.98. This calculation is beyond the scope of this course, so a pH of 7 is a good approximation.

15D.5. What is the difference between (a) monoprotic acid and a diprotic acid, (b) a weak acid and a strong acid, (c) an acid and a base?

What does it mean if an acid is a) strong? b) polyprotic?

a) The acid completely ionizes in water; every single molecule of acid releases at least one proton. b) Each molecule of a polyprotic acid can (doesn’t always) release more than one proton into solution.

You dissolve 50. g of HCl in water to make 2.0 liters of solution. You then take 1000. mL of the acid and decide to neutralize it with Ca(OH)2; how many grams of Ca(OH)2 must you use?

50 g HCl

HCl, HBr, and HI are strong acids, yet HF is a weak acid. What does this mean in terms of the extent to which these substances are ionized in solution? Label each of the following substances as an acid, base, salt, or none of the above. Indicate whether the substance exists in aqueous solution entirely in molecular form, entirely as ions, or as a mixture of molecules and ions. (a) HF; (b) acetonitrile, CH3CN; (c) NaClO4; (d) Ba(OH)2. 15D.6. Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resultant solution. The sodium bicarbonate reacts with sulfuric acid as follows: 2 NaHCO3(s) + H2SO4(aq) Na2SO4(aq) + 2H2O(l) + 2CO2(g) Sodium bicarbonate is added until the fizzing due to the formation of CO2(g) stops. If 27 mL of 6.0 M H2SO4 was spilled, what is the minimum mass of NaHCO3 that must be added to the spill to neutralize the acid?

1 mol 1 = .686 M 36.46 g HCl 2.0 L

2 HCl + Ca (OH ) 2  → 2 H 2 O + Ca 2 + + 2Cl − .686 mol HCl 1 mol Ca( OH ) 2 74.09 g = L 2 mol HCl 1 mol Ca (OH ) 2 25 g Ca(OH)2

Practice Sheet 15 Answers 15D.1

a) b)

 1 mol HNO 3 6.3 g HNO3 ×   63.01 g HNO3

 1 mol H + ion  1   = 0.2 M H+ → pH = 0.70   1 mol HNO . 500 L 3   

H2SO4 is able to ionize twice because it is diprotic (it has 2 H+ ions). The first ionization is strong, meaning that almost all of H2SO4 will dissociate into hydronium ion (H30+) and bisulfate ion (HSO4-). However, the second ionization is weak, meaning that only some of the bisulfate is going to dissociate into hydronium ion and sulfate ion (SO42-). To find the pH range, first find the pH assuming that none of the bisulfate dissociates. This will be your maximum pH. Then, to find your minimum pH calculate the pH assuming that both the H2SO4 and bisulfate (HSO4-) completely dissociates. These two pH values are your range.

 1 mol H 2 SO 4   1 mol H + ion   1      9.8 g H2SO4 ×   = 0.2 M H+ → pH = .70 98 . 0 g H SO 1 mol H SO 0 . 500 L  2 4  2 4  

 1 mol H 2 SO 4 9.8 g H2SO4 ×   98.0 g H 2 SO4

  2 mol H + ion   1     = 0.4 M H+ → pH = .40     1 mol H 2 SO4   0.500 L 

The pH will be between .70 and .40. 15D.2 [H+] = 10pH = 10-2.34 = 0.00457 M H+. Since there is a 1:1 ration of H+ released to molecules of acid, this is also the concentration of the acid = 0.0046 M

15D.3 a) pH = 11.89 so pOH = 14 -11.89 = 2.11 so [OH] = 10-2.11 = 0.00776 M, this is also the concentration of KOH. b) pH = 11.68 so pOH = 14-11.68 = 2.32 so [OH] = 10-2.32 = .00479 M  1 mol Ca( OH ) 2   = .00239 M Ca(OH)2 Then, .00479 M OH- ×  − 2 mol OH  

15D.4 Your lab partner may seem tricky, but there is a flaw in her logic. For very dilute solutions like this one, we cannot neglect the autoprotolysis of water: 2H2O = H3O+ + OH–. Remember that in pure water, [H3O+] = [OH–] = 1×10–7 M, which is larger than the 2.0×10–9 M HI solution. To correctly solve this problem, we would need to account for the ionization of water simultaneously with the dissociation of the HI, which would give a pH in the acidic range.

15D.5 a) A monoprotic acid has one ionizable H and a diprotic acid has two. b) A strong acid is completely ionized in aqueous solution, whereas only a portion of the molecules are ionized in a weak acid solution. c) An acid gives up protons in solution ( ↑H+) while a base accepts them( ↑OH–).

The strong acid will exist entirely as ions while the weak acid will reach some sort of equilibrium between ions and neutral molecules. There are at least as many molecules as there are anions or cations. a) HF: weak acid, mixture of molecules and ions b) CH3CN: none of the above, entirely molecules c) NaClO4: salt, entirely ions d) Ba(OH)2: base/salt, entirely ions 15D.6

2NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) + 2CO2(g)

 6.0 mol H2 SO4 0.027 L H2SO4 ×  L 

  2 mol NaHCO3     1 mol H 2SO 4

  84.01 g NaHCO3   = 27 g NaHCO3   1 mol NaHCO 3  ...