CHhmwk5 - moore(jwm2685 – Homework 5 – vanden bout...

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Unformatted text preview: moore (jwm2685) – Homework 5 – vanden bout – (51640) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. No problems in Homework 5 require next week’s lecture to be solved. 001 10.0 points A 50.0 mL sample of Ca(OH) 2 is neutralized by 300.0 mL of HCl solution with a pH of 1.3. Calculate the molarity of the Ca(OH) 2 solution. 1. 0.05 M 2. 1.0 M 3. 0.30 M 4. 0.50 M 5. 0.15 M correct Explanation: First, calculate the H + concentration by using inverse log. M H + = 0.05012. Then, calculate M of the base: . 05012 M HCl × . 300 L × 1 mol HCl 2 mol Ca(OH) 2 × 1 . 0500 L = 0 . 15036 = 0 . 15 M . 002 10.0 points A 200 mL sample of Ba(OH 2 ) is titrated to pH 7 with 300 mL of 0 . 053 M HClO 4 . What was the pH of the original Ba(OH 2 ) solution, before any HClO 4 was added? 1. 13 . 2 2. 1 . 1 3. 12 . 9 correct 4. . 796 Explanation: Because the problem stipulates that the sam- ple was titrated to pH 7, we know that the moles of hydroxide ions in the analyte were equal to the number of moles of protons added by the titrant. n OH- = n H + V Ba(OH) 2 · [OH- ] = V HClO 4 · [H + ] [OH- ] = V HClO 4 V Ba(OH) 2 [H + ] = 300 200 . 053 = 0 . 08 pOH = 1 . 1 pH = 12 . 9 003 (part 1 of 5) 10.0 points Consider the following titration curve. 2 4 6 8 10 12 14 20 40 60 80 100 120 140 Volume of base (mL) pH This is a plot of a (weak, strong) acid and a (weak, strong) base titration. 1. strong; strong 2. strong; weak 3. weak; strong correct Explanation: The curve shape follows that for a weak acid titrated with a strong base, identified by the concave down-to-concave up shape of the curve before the equivalence point. 004 (part 2 of 5) 10.0 points What is the approximate p K a of the acid? moore (jwm2685) – Homework 5 – vanden bout – (51640) 2 1. 10 2. 5 3. 13 4. 7.5 correct 5. 12.5 6. There is no p K a because it is a strong acid. Explanation: The p K a of the strong acid can be estimated by inspection of the buffer region, or by the pH at half the equivalence point volume. The equivalence point is at 70 mL added base, so the pKa is at 35 mL, about pH=7.5. 005 (part 3 of 5) 10.0 points What is the pH at the equivalence (stoichio- metric) point? 1. about 10 correct 2. exactly 7 3. about 7.5 4. about 5 Explanation: The equivalence point is the vertical part of the curve, which is for this curve the second inflection point (changes from concave up to concave down) at pH = 10. 006 (part 4 of 5) 10.0 points What type of salt is produced as a result of this acid/base reaction? 1. basic correct 2. acidic 3. neutral Explanation: The salt is formed from the neutralization of a weak acid and a strong base, so it is a basic salt....
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CHhmwk5 - moore(jwm2685 – Homework 5 – vanden bout...

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