# Acids, Bases and Buffers

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Acids, Bases and Buffers • • • • • • • • • • • • • • • • • • • • • •

Strong vs weak acids and bases Equilibrium as it relates to acids and bases pH scale: [H+(aq)] to pH, pOH, etc pH of weak acids pH of strong acids Conceptual about oxides (for multiple choice) metal vs nonmetal oxides polyprotic acids and their successive Ka’s Definitions of acids and bases: Lewis, Arrenhius, Bronstead Lowry Conjugate acids ands bases Anions of weak acids are bases Cations of weak bases are acids Buffers: Weak acid plus its conjugate base (the common ion) Buffers: Weak bases plus conjugate acid (the common ion) Calculating the pH of a buffered solution. Henderson-Hasselbach equation as a shortcut for finding the pH of a buffered solution only Titration's: Calculate pH and other things given a titration of acids and bases. Ignore x assumption Make sure to watch your sig figs Show all equilibrium expressions for full credit Initial change and final moles reacting. Use good limiting reagent skills when solving these problems Ksp and solving these problems Common ion and Ksp: Think about LeChatleier

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12 X

10

X

8

X X

pH 6 4X

X

X

X

2 0 0

5

10 15 20 25 millilitres of NaOH

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#1. A 30.00 millilitre sample of a weak monoprotic acid was titrated with a standardized solution of NaOH. A pH meter was used to measure the pH after each increment of NaOH was added, and the curve above was constructed. (a) Explain how this curve could be used to determine the molarity of the acid. (b) Explain how this curve could be used to determine the dissociation constant Ka of the weak monoprotic acid. (c) If you were to repeat the titration using a indicator in the acid to signal the endpoint, which of the following indicators should you select? Give the reason for your choice. Methyl red Ka = 1×10–5 Cresol red Ka = 1×10–8 Alizarin yellow Ka = 1×10–11 (d) Sketch the titration curve that would result if the weak monoprotic acid were replaced by a strong monoprotic acid, such as HCl of the same molarity. Identify differences between this titration curve and the curve shown above.

#2 In an experiment to determine the molecular weight and the ionization constant for ascorbic acid (vitamin C), a student dissolved 1.3717 grams of the acid in water to make 50.00 millilitres of solution. The entire solution was titrated with a 0.2211 molar NaOH solution. The pH was monitored throughout the titration. The equivalence point was reached when 35.23 millilitres of the base has been added. Under the conditions of this experiment, ascorbic acid acts as a monoprotic acid that can be represented as HA. (a) From the information above, calculate the molecular weight of ascorbic acid. (b) When 20.00 millilitres of NaOH had been added during the titration, the pH of the solution was 4.23. Calculate the acid ionization constant for ascorbic acid. (c) Calculate the equilibrium constant for the reaction of the ascorbate ion, A–, with water. (d) Calculate the pH of the solution at the equivalence point of the titration.

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#3 Give a brief explanation for each of the following. (a) For the diprotic acid H2S, the first dissociation constant is larger than the second dissociation constant by about 105 (K1 ≈ 105K2). (b) In water, NaOH is a base but HOCl is an acid. (c) HCl and HI are equally strong acids in water but, in pure acetic acid, HI is a stronger acid than HCl. (d) When each is dissolved in water, HCl is a much stronger acid than HF.

#4 The acid ionization constant, Ka, for propanoic acid, C2H5COOH, is 1.3×10–5. (a) Calculate the hydrogen ion concentration, [H+], in a 0.20–molar solution of propanoic acid. (b) Calculate the percentage of propanoic acid molecules that are ionized in the solution in (a). (c) What is the ratio of the concentration of propanoate ion, C2H5COO–, to that of propanoic acid in a buffer solution with a pH of 5.20? (d) In a 100.–milliliter sample of a different buffer solution, the propanoic acid concentration is 0.35–molar and the sodium propanoate concentration is 0.50–molar. To this buffer solution, 0.0040 mole of solid NaOH is added. Calculate the pH of the resulting solution.

#5 The equations and constants for the dissociation of three different acids are given below. HCO3– ---------> H+ + CO32– Ka = 4.2 × 10–7 H2PO4– ----------> H+ + HPO42–

Ka = 6.2 × 10–8

HSO4– ---------> H+ + SO42– Ka = 1.3 × 10–2 (a) From the systems above, identify the conjugate pair that is best for preparing a buffer with a pH of 7.2. Explain your choice. (b) Explain briefly how you would prepare the buffer solution described in (a) with the conjugate pair you have chosen. (c) If the concentrations of both the acid and the conjugate base you have chosen were doubled, how would the pH be affected? Explain how the capacity of the buffer is affected by this change in concentrations of acid and base. (d) Explain briefly how you could prepare the buffer solution in (a) if you had available the solid salt of the only one member of the conjugate pair and solution of a strong acid and a strong base.

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#6 CH3NH2 + H2O - CH3NH3+ + OH– Methylamine, CH3NH2, is a weak base that reacts according to the equation above. The value of the ionization constant, Kb, is 5.25×10–4. Methylamine forms salts such as methylammonium nitrate, (CH3NH3+)(NO3–). (a) Calculate the hydroxide ion concentration, [OH–], of a 0.225–molar aqueous solution of methylamine. (b) Calculate the pH of a solution made by adding 0.0100 mole of solid methylammonium nitrate to 120.0 milliliters of a 0.225–molar solution of methylamine. Assume no volume change occurs. (c) How many moles of either NaOH or HCl (state clearly which you choose) should be added to the solution in (b) to produce a solution that has a pH of 11.00? Assume that no volume change occurs. (d) A volume of 100. milliliters of distilled water is added to the solution in (c). How is the pH of the solution affected? Explain.

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pH

0

mL HCl (d) Write the equation for the reaction that occurs if a few additional milliliters of the HCl solution are added to the solution resulting from the titration in (c).

#9: HOBr (aq)   H+(aq) + OBr-(aq)

Ka = 2.3 x 10-9

Hypobromous acid, HOBr, is a weak acid that dissociates in water as represented by the equation above. a) Calculate the value of [H+] in a an HOBr solution that has a pH of 4.95 b) Write the equilibrium constant expression for the ionization of HOBr in water, then calculate the concentration of HOBr in an HOBr solution that has [H+] equal to 1.8 x 10-5 M. c) A solution of Ba(OH)2 is titrated into a soluton of HOBr. Calculate the volume of 0.115 M Ba(OH)2 needed to reach the i) equivalence point when titrated into 65.0 mL sample of 0.146 M HOBr. ii) Indicate wheter the pH at the equivalence point is less than 7, equal to 7, greater then 7. Explain. d) Calcluate the number of moles of NaOBr(s) that would have to be added to 125 mL of 0.160 M HOBr to produce a buffer solution with [H+] = 5.00 x 10-9 M. Assume that volume change is negligible. e) HOBr is a weaker acid than HBrO3. Account for this fact in terms of molecular structure.

#10: Ka = [H3O+][OCl-] [HOCl]

= 3.2 X 10-8

Hypochlorous acid, HOCl, is a weak acid in water. The Ka expression for HOCl is shown above. (a)

Write a chemical equation showing how HOCl behaves as an acid in water.

6 (b) (c) (d)

Calculate the pH of a 0.175 M solution of HOCl. Write the net ionic equation for the reaction between the weak acid HOCl(aq) and the strong base NaOH(aq). In an experiment, 20.0 mL of 0.175 M HOCl(aq) is placed in a flask and titrated with 6.55mL of 0.435 M NaOH(aq). (i) Calculate the number of moles of NaOH(aq) added. (ii) Calculate the [H3O+] in the flask after the NaOH(aq) has been added. (iii) Calculate [OH-] in the flask

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ANSWERS #1 Answer (a) The sharp vertical rise in pH on the pH–volume curve appears at the equivalence point (about 23 mL). Because the acid is monoprotic, the number of moles of acid equals the number of moles of NaOH. That number is the product of the exact volume and the molarity of the NaOH. The molarity of the acid is the number of moles of the acid divided by 0.30L, the volume of the acid. (b) At the half–equivalence point (where the volume of the base added is exactly half its volume at the equivalence point), the concentration [HX] of the weak acid equals the concentration [X–] of its anion. Thus, in the equilibrium expression [H+][X–]/[HX] = Ka, [H+] = Ka. Therefore, pH at the half–equivalence point equals pKa. (c) Cresol red is the best indicator because its pKa (about 8) appears midway in the steep equivalence region. This insures that at the equivalence point the maximum color change for the minimal change in the volume of NaOH added is observed.

(d) 12 10 8 pH 6

Longer equivalence region for strong acid

X X X X

X X X volume of NaOH 4Same X added for equivalence point 2 Acid portion at lower 0 pH for strong acid 0 5 10 15 20 25 30 millilitres of NaOH

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#2 Answer (a) (0.2211M)(0.03523L) = 7.789×10–3 mol 1.3717g/7.789×10–3 mol = 176.1g/mol (b) at pH 4.23, [H+] = 8.0×10–8M (0.02000L)(0.2211mol ⋅ L−1 ) [A ] = = 0.06317M 0.07000L (0.01523L)(0.2211mol ⋅ L−1 ) [HA] = = 0.04810M 0.07000L [H + ][A − ] (5.9 × 10 −5 )(0.06317) −5 K= = = 7.7 × 10 [HA] (0.04810) (c) A– + H2O ⁄ HA + OH– −

K=

[OH − ][HA] [A − ]

=

Kw 1.×10 −14 = = 1.3 × 10 −10 −5 Ka 7.7 × 10

(d) at equiv. pt.

7.789 × 10 −3 mol [A ] = = 0.0914M 0.08523L [OH–]2 = (1.3×10–10)(9.14×10–2) = 1.2×10–11 [OH–] = 3.4×10–6M pOH =–log(3.4×10–6) =5.47; pH = (14–5.47)= 8.53 −

#3 Answer: (a) After the first H+ is lost from H2S, the remaining species, HS–, has a negative charge. This increases the attraction of the S atom for the bonding electrons in HS–. Therefore, the bond is stronger, H+ is harder to remove, and K2 is lower. (b) Polar H2O can separate ionic NaOH into Na+(aq) and OH–(aq), giving a basic solution. In HOCl, chlorine has a high attraction for electrons due to its greater charge density. This draws electrons in the H–O bond towards it and weakens the bond. H+ can be removed, making an acid solution. (c) Water is a more basic solvent (greater attraction for H+) and removes H+ from HCl and HI equally. Acetic acid has little attraction for H+, but the H+ separates from the larger I– more easily than from the smaller Cl–. (d) The bond between H and Cl is weaker than the bond between H and F. Therefore, HCl is a stronger acid.

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#4 Answer: [H+][C2H5COO–] (a) = Ka [C2H5COOH] [H+] = [C2H5COO–] = X [C2H5COOH] = 0.20 M – X assume X is small, ∴, 0.20 – X ≈ 0.20 X2 −5 −3 + = 1.3×10 ;X =1.6×10 M = [H ] 0.20 (b) from (a), X = amount of acid that ionized, ∴, 1.6∞10–3 × 100% = 0.81% ionized 0.20 (c) @ pH 5.20, [H+] = antilog (–5.20) = 6.31×10–6 M (6.3×10−6 )[C2H 5COO − ] −5 = K a = 1.3×10 [C2H 5COOH] [C2H5COO–] 2.1 = 1 [C2H5COOH]

OR (c) Henderson–Hasselbalch [base] pH = pK a + log [acid] -

[C H O ] 5.20 = 4.89 + log 3 5 2 [HC 3H 5O2] -

[C H O ] log 3 5 2 = 0.31 = 2.1 [HC 3H 5O2] (d) [C2H5COO–] = 0.50 M; [C2H5COOH] = 0.35 M [OH–] = 0.0040 mol/0.100 L = 0.040 M this neutralizes 0.04 M of the acid, giving [C2H5COOH] = 0.31 M and the propanoate ion increases by a similar amount to 0.54 M. +

[H ](0.54) -5 + -6 = 1.3 x 10 , [H ] = 7.5 x 10 M 0.31 pH = – log [H+] = 5.13 OR (d) using [ ]’s or moles of propanoic acid and propanoate ion... 0.54 pH = pK a + log 0.31 = 4.89 + 0.24 = 5.13

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#5 Answer: (a) Best conjugate pair: H2PO4–, HPO42–. When 7.2 = pH = pKa for this pair when [H2PO4–] = [HPO42–]. (b) Dissolve equal moles (or amounts) of H2PO4–, and HPO42– (or appropiate compounds) in water. (c) pH not changed. Capacity of buffer would increase because there are more moles of conjugate acid and conjugate base to react with added base or acid. (d) Add strong base to salt of conjugate acid OR add strong acid to salt of conjugate base. Add 1 mole conjugate acid to 1/2 mole strong base OR 1 mole conjugate base to 1/2 mole strong acid. OR Use pH meter to monitor addition of strong base to conjugate acid OR strong acid to conjugate base.

#6 Answer: [CH 3NH +3 ][OH − ] (a) K b = [CH 3NH 2 ] CH3NH2 + H2O ⁄ CH3NH3+ + OH– [ ]i 0.225 0 0 ∆[ ] –X +X +X X X [ ]eq 0.225–X

Kb = 5.25×10−4 =

[X][X] X2 ≅ [0.225 − X] 0.225

X = [OH–] = 1.09×10–2 M solved using quadratic: X = [OH–] = 1.06×10–2 M (b) [CH3NH3+] = 0.0100 mol / 0.1200 L = 0.0833 M or CH3NH2 = 0.120 L × 0.225 mol/L = 0.0270 mol [0.0833 + X][X] 0.0833X −4 K b = 5.25×10 = ≅ [0.225 − X] 0.225

X = [OH–] = 1.42×10–3 M; pOH = 2.85; pH = 11.15 OR

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pH = pKa + log Ka =

[base] [acid]

1×10−14 5.25×10

−4

pH = 10.72 + log

= 1.91×10− 11; pKa =10.72 (0.225) =11.15 (0.0833)

OR

[acid] ; pKb = 3.28 [base] (0.0833) pOH = 3.28 + log = 2.85; pH =11.15 (0.225) (c) HCl must be added. [0.0833 + X][0.0010] −4 Kb = 5.25×10 = [0.225 − X] X = 0.0228 M 0.0228 mol/L × 0.120 L = 2.74×10–3 mol HCl OR [base] [base] 11.00 = 10.72 + log ; log = 0.28 [acid] [acid] [base] (0.225 − X) =1.905 = ; X = 0.0227M [acid] (0.0833 + x) pOH = pKb + log

0.0227 M × 0.120 L = 2.73×10–3 mol HCl +

[CH3NH3 ] (d) The ratio does not change in this buffer solution with dilution, therefore, [CH3NH2] no effect on pH.

#7 Answer: (a) Zn is oxidized to Zn2+ by H+ which in turn is reduced by Zn to H2. Identify H2(g) or Zn dissolving as Zn2+. Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g) Explicit: Redox or e– transfer or correctly identify oxidizing agent or reducing agent. (b) H2SO4 dissociates, forms ions or dydration “event”. Bonds form, therefore, energy given off (connection). (c) BaSO4 (ppt) forms or H+ + OH– form water. Newly formed water and ppt remove ions lowering conductivity. 2+ Ba (aq) + OH–(aq) + H+(aq) + SO42–(aq) → BaSO4(s)+ H2O(l) (d) First 10 mL produces solution of SO42– and OH– or excess OH– or partial

12 neutralization (pH 13.0 → 12.6). [presence of HSO4– in solution voids this point] Second 10 mL produces equivalence where pH decreases (changes) rapidly (pH 12.6 → 7.0). [pH “rises” or wrong graph, if used, voids this point]

#8 Answer: (a) PO43– + H+ → HPO42–; HPO42– + H+ → H2PO4– (b) HPO42– . .– : . .:. . .. O O . . ::.P. : O . .: H :O . .: – (c)

3–

2–

PO4+ H +→ HPO4 pH

0 (d)

HPO42–+ H+→ H2PO4 mL HCl

H+ + H2PO4– → H3PO4

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pH = −log[H + ] 10− pH = [H + ] 10−4.95 = 1.12x10−5 M HOBr ↔ H + + OBr− Ka = b.

[H + ][OBr − ] = 2.3x10−9 [HOBr]

[1.8x10−5 ][1.8x10−5 ] 2.3x10 = [HOBr] [HOBr] = 0.14 M −9

14 c. 2HOBr + Ba(OH) 2 → Ba(OBr) 2 + 2HOH

65.0mL 0.146M

0.115M

0.146molHOBr 0.0650LHOBr 1molBa(OH) 2 1LBa(OH) 2 x x x = 1LHOBr 1 2molHOBr 0.115molBa(OH) 2 0.0412L = 41.2mLBa(OH) 2 pH at the equivalence point will be greater than 7. Here we have a weak acid reacting with a strong base. At the equivalence point we will have the OBr- ion and that ion is a base: Obr- + HOH  OH- + HOBr d. bufferpH = −log(5.00 x10−9 ) = 8.30

pKa = −log Ka = −log(2.3 x10−9 ) = 8.64 MV = moles : (0.160 M )(0.125 L) = 0.02 moles pH = pKa + log

[OBr− ] or [ HOBr]

[ molesOBr− ] pH = pKa + log [ molesHOBr]

x 0.020 moles fromSolver : x = 0.0091moles

8.30 = 8.64 + log

e. HOBr is a weaker acid than HBrO3 because of the extra highly electronegative oxygens on the HBrO3. The more electronegative oxygen atoms draws electons to the side of the atom with the oxygen. This makes the molecule more polar and thus makes it more soluble in water and have a greater propensity to dissociate. Stronger acids dissociate more and since this is more polar it will dissociate more.

## Acids, Bases and Buffers

1 Acids, Bases and Buffers • • • • • • • • • • • • • • • • • • • • • • Strong vs weak acids and bases Equilibrium as it relates to acids and bases p...

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