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# 019 34points the un ionized form of an acid indicator

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019 3.4points The un-ionized form of an acid indicator is yellow and its anion is blue. The K a of this indicator is 10 - 5 . What will be the color of the indicator in a solution of pH 3? 1. yellow correct 2. red 3. green 4. blue 5. orange Explanation: K a = 10 - 5 pH = 3 HIn yellow + H 2 O H 3 O + + In - blue K a = 10 - 5 p K a = - log ( 10 - 5 ) = 5 The color change range is pH = p K a ± 1. At pH values above 6 the indicator will be ionized and at pH values below 4 the indicator will be un-ionized. 020 3.4points Which of the following acid-base indicators should be used for the titration of NH 3 with HBr? K b of ammonia is 1.8 × 10 - 5 . 1. Neutral red, color change red/yellow 6 . 8 < pH < 8 . 0 2. Methyl red, color change red/yellow at 4 . 4 < pH < 6 . 2 correct 3. None of these is suitable. 4. Any of these is suitable. 5. Phenolphthalein, color change colorless/red-violet 8 . 0 < pH < 10 . 0 Explanation: The endpoint of a weak base (NH 3 ) - strong acid (HBr) titration is below pH 7. An appro- priate indicator has color range pH = p K a ± 1.

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casey (rmc2555) – Homework 9 – holcombe – (51395) 7 021 3.4points The curve for the titration of hydrofluoric acid (HF) with NaOH(aq) base is given below. 0 2 4 6 8 10 12 14 0 20 40 60 80 100 120 140 Volume of base (mL) pH Estimate the p K a of hydrofluoric acid. C a = 0 . 64, C b = 0 . 48, and the volume of HF is 100 mL. 1. 37 . 5 2. 75 3. 3 . 1 correct 4. 8 . 07 5. None of these Explanation: K a = 7 . 2 × 10 - 4 C a = 0 . 64 C b = 0 . 48 V HF = 100 mL 0 2 4 6 8 10 12 14 0 20 40 60 80 100 120 140 Volume of base (mL) pH (75,8 . 07) (37 . 5,3 . 1) The equivalence point of this titration is when the curve is at an inflection point (nearly vertical); i.e. , at a volume of 75 mL . The pH at the equivalence point of this titration is 8 . 07 pH . The p K a can be found at one-half the vol- ume of the equivalence point; i.e. , at 37 . 5 mL. The p K a is 3 . 1 pH from looking at the graph. The formula is p K a = - log ( K a ) = - log parenleftBig 7 . 2 × 10 - 4 parenrightBig = 3 . 14267 pH . Note : The p K a is the pH when the mole fraction is 0.5. 022 3.4points At the stoichiometric point in the titration of 0.260 M CH 3 NH 2 (aq) with 0.260 M HCl(aq), 1. [CH 3 NH + 3 ] = 0.260 M. 2. the pH is less than 7. correct 3. the pH is greater than 7. 4. [CH 3 NH 2 ] = 0.130 M. 5. the pH is 7.0. Explanation:
casey (rmc2555) – Homework 9 – holcombe – (51395) 8 023 3.4points Calculate the pH at the equivalence point for the titration of a solution containing 150.0 mg of ethylamine (C 2 H 5 NH 2 ) with 0.1000 M HCl solution. The volume of the solution at the equivalence point is 250.0 mL. K b for ethylamine is 4 . 7 × 10 - 4 . Correct answer: 6 . 27392. Explanation: m C 2 H 5 NH 2 = 150.0 mg [HCl] =0 . 1000 M V = 250.0 mL K b = 4 . 7 × 10 - 4 C 2 H 5 NH 2 + HCl C 2 H 5 NH + 3 + Cl - At the equivalence point all of the C 2 H 5 NH 2 has reacted and no excess HCl remains. mol C 2 H 5 NH + 3 formed = 0 . 150 45 . 07 = 3 . 32816 × 10 - 3 [CH 5 NH + 3 ] = 3 . 32816 × 10 - 3 mol 250 × 10 - 3 L = 1 . 33126 × 10 - 2 M . The CH 3 NH + 3 hydrolyzes. CH 3 NH + 3 +H 2 O CH 3 NH 2 + H 3 O + ini 0 . 0133126 0 0 Δ - x x x fin 0 . 0133126 - x x x K a = K w K b = [CH 3 NH 2 ][H 3 O + ] [CH 3 NH + 3 ] 1 . 0 × 10 - 14 4 . 7 × 10 - 4 = x 2 1 . 33126 × 10 - 2 - x x 2 1 . 33126 × 10 - 2 x = H 3 O + ] = 5 . 3221 × 10 - 7 assumption valid pH = 6 . 27392 024 3.4points The figure below shows the titration of a monoprotic weak acid with a strong base.

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• Fall '07
• Holcombe

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