Homework Chapter 18 K & T Acids and Bases

11. 0.050L of 0.40M NH_3 + 0.050L of 0.40M HCl. What concentrations?

1. moles NH_3 = 0.050*0.40 = 0.020 mole
2. moles HCl = 0.020 mole
3. total volume = 100 mL = 0.10 L
4. initial concentrations = 0.020 mol/0.10L = 0.2M

What is the reaction?

With such a large K "all" the NH₃ will become NH₄⁺ and "all " the H₃O⁺ will be consummed. We then have the dissociation of NH₄⁺ water acting as a base.

Use the tabular scheme:

17. Aniline is and will be designated by .
The conjugate acid dissociation is given in the text as and from the relationship Kb = Kw/Ka we get the equilibrium constant for the base reaction
.

The equivalence point is where the base, , is completely neutralized by the added acid. That means the base is gone, has zero concentration, is converted to the conjugate acid,. This acid may hydrolyze water to reform an amount of base so that the equilibrium constant expression is satisfied.

1. We need to know the moles of acid added for neutralization of the base.
2. Moles acid = 0.02567 L x 0.0175 M = 0.0045 moles.
3. The total volume = 25 mL + 25.67 mL = 0.05067L
4. The concentration of is 0.0045mol/0.05067L = 0.0887M

Use the tabular method.

Pluging into the equilibrium constant expression: .

1. Since Ka is small x must be small.
2. Assume x is small with respect to 0.089M and solve the resulting approximate equation to find x = 1.46 x 10^{-3}M = [H3O^+]. This is reasonably small w.r.to 0.089M.
3. pH = 2.84
4. pOH = 14 - 2,84 = 11.16, [OH^-] = 6.85 x 10^{-12}M. and
5. since x is so small [] = 0.089M.
6. This all came from . Moles = 0.089M*0.05067L = 0.0045 moles
7. The original volume was 25mL and so the orignial concentration is [] = 0.0045/0.025 = 0.18 M.

19.
(a) Effect: add

Result: Lower pH

is a weak base with (see value for Ka in problem 23 chap 17 page 842 and Kb = Kw/Ka). The equilibrium is governed by . As we add , is increased producing both OH^- and driving left removing H^+. Therefore, get lower pH.

25.
2.75 g of lactate added to 500 mL aqueous solution that is 0.1M in lactic acid. What is the pH? Is this a buffer? First, estimate concentration of . 2.75g/113.1 g/mol (sodium lactate-don't forget the atomic wt. of Na) = 0.024 mole lactate in 0.5L or 0.048M in .

; and

Looks to be in the buffer region and the pH is lower than the pH of the lactic acid solution.

27. pH = pKa - Log[B]/[A] = 4.5 = 4.75 - Log[B]/[A]; Log[B]/[A] = 0.25; [B]/[A] = 1.8; [A] = 0.1M so [B] = 0.18M. 1L = 0.18 moles of acetate. Mw sodium acetate = 82 g/mol and therefore 14.7 g sodium acetate added to 1L.

37.

and

39.
(a) Yet another mickey mouse problem.

(b) 9.15 - 0.5 = 8.65 = 9.25 + Log(B/A); Log(B/A) = -0.60 and [NH_3]/[NH_4^+] = 1/4

41. A titration curve has four simple regions.

• First, the starting pH. In the case of pyridine we have a base and the pH > 7
• Second, the region where base and conjugate acid are equal in concentratin (the buffer region) has pH = pKb also greater than pH = 7. For pyradine pKb = 8.82.
• Third, when the base has been completely neutralized by the acid (the equivalence point) we can consider all the base is the conjugate acid. This conjugate acid hydrolyzes water to form an acidic solution. A^+ + H_2O -> HOA + H^+
• Fourth, when excess acid is added and the pH is just due to excess acid in an increasing volume.
  

• Solution volume at equivalence point? Total moles pyridine = 0.050L x 0.050M = 0.0025 moles. Need 0.0025 moles of acid to neutralize this base. Conc of HCl = 0.10M therefore need 0.0025/0.10 = 0.025 L = 25 mL. Total volume = 50 mL + 25 mL = 75 mL.

  42. 25 mL 0.11M titrated with 0.10M HCl.
  (a) The pH of the starting solution is just the pH of the 25mL of 0.11M ammonia.
  

  x = [OH^-] = 0.0014; pOH = 2.85; pH = 14 - pOH = 11.15.

  (b) pH at equivalence point. The equivalence pt is where the base, in this case, is completely neutralized. That is, it is all gone. The concentration of that is formed is equal to the original concentration of ammonia modified by the addition of a certain volume of HCl. The moles of ammonia = 0.11M x 0.025L = 0.00275 moles. This requires an equal number of moles of HCl. Since the HCl concentration is 0.1M we need 0.00275 moles/0.1M = 0.0275L of HCl. The total volume of solution is now 0.025L + 0.0275L = 0.0525L = 52.5mL. The concentration of is 0.00275 moles/0.0525L = 0.0524M. This is the initial concentration of that ionizes to form an acidic solution. Now to the tabular method.

  

  From this using the usual approximation neglecting x with respect to 0.0524 we get pH = 5.3.

  (c) The midpoint of the titration is the half neutralization point, or the "buffer" point. Using the equilibrium constant expression for the we get

  and at half neutralization [NH_4^+] = [NH_3] and pOH = pKb = 4.74 and pH = 14 - 4.74 = 9.3.

  (d) An indicator that changes color at pH = 5.3 is best.

  (e) The indicator is bromcresol green.

  The acid neutralizes the base. The following table considers the volumes of acid added computing the ammonia concentration. The last line is where all of the ammonia has been neutralized and excess acid added.

  

  51. 0.515 g phenol in 100mL water. 0.123M OH^- is used in a titration. What pH and concentrations of ions at equivalence point?
  

  The molarity of the phenol in 100 mL of water = (0.515/94.1)/0.1L = 0.0547M. (0.515/94.1) = 0.00547 moles phenol requires 0.00547 moles/0.123M = 0.0445L of hydroxide. Total volume = 100mL + 44.5mL = 144.5mL. Concentration of the base = 0.00547/0.1445 = 0.0379M. and

  

  [OH^-] = [] = 0.0015M, pH = 11.18. [] = 0.0379 - 0.0015 = 0.0364M. The sodium concentration = no. moles OH^ added/total vol = 0.00547 moles/0.1445 = 0.0379M

  The titration curve for phenol with OH^-:

  

  56. Buffer Capacity = Change pH of buffer by one pH unit. This is an effort to measure the control that a buffer exerts. The expression relating pH and pKa to the acid and base concentrations is:
  

  The acetic acid concentration = acetate concentration = 0.1M and were we to add x moles/L of OH^- the pH would change according to:

  from which we get x = 0.9/11 = 0.082 moles per liter of base added. This is a pretty significant fraction of the acid and conjugate weak base present.

  63. Mw trimethylamine = 59.1 g/mole, 0.221 g -> 0.00374 moles trimethylamine. Initial concentration is 0.00374/0.050 = 0.0748M.

  (a) The pH of the starting solution:
  

  [OH^-] = x = 0.00235, pH = 11.4

  (b) At the midpoint, or half neutralization point, or the mid-buffer region, the pH = pKb = 4.13.

  (c) At the equivalence point the number of moles of the conjugate acid = original moles of the base = 0.00374 moles. Required this number of moles of HCl for neutralizatoin and at 0.1M need 0.00374/0.1 = 0.0374 L. Total volume is now 37.4mL + 25mL = 0.0624L. The concentration of conjugate acid = 0.00374/0.0624 = 0.060M.

  

  [H^+] = x = 2.847 x 10^{-6} or pH = 5.55.

  (d)
  

  67. Titrations identified as follows:

  (c) (d)

  (a)

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