Make up 2-solutions - Version 139 Make up 2 laude (51635)...

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Version 139 – Make up 2 – laude – (51635) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 6.0 points Hint: You need to do the RICE diagram For this one. The ratio oF acid to base is not 1 to 1. What is the fnal pH oF a solution containing 100 mL oF 0.1 M HX and 400 mL oF 0.2 M NaX aFter 0.02 mol oF HNO 3 is added? The p K a is 3.00. 1. 3.30 correct 2. 2.80 3. 6.40 4. 2.20 5. 3.00 Explanation: Initially (100 mL) (0 . 1 M) = 10 mmol HA (400 mL) (0 . 2 M) = 80 mmol A - Now add the impurity: 0.02 mol oF HNO 3 = 20 mmol H + : A - + H + HA ini 80 0 10 Δ - 20 +20 fn 60 30 Thus A - HA = 60 30 = 2 and pH = 3 . 0 + log(2) = 3 . 30103 . 002 6.0 points Consider the aqueous solution initially with H 3 PO 4 , NH 3 , and NH 4 H 2 PO 4 . A system oF equations is needed to determine the equilib- rium concentrations. Which oF the Following equations correctly balances the charge For the system? 1. [H + ] = [OH - ] 2. [NH 4 + ] = [H 2 PO 4 - ] + 2[HPO 4 2 - ] + 3[PO 4 3 - ] 3. [H + ] + [NH 4 + ] = [OH - ] + [H 2 PO 4 - ] + [HPO 4 2 - ] + [PO 4 3 - ] 4. [H + ] + [NH 4 + ] = [OH - ] + [H 2 PO 4 - ] 5. [H + ] + [NH 4 + ] = [OH - ] + [H 2 PO 4 - ] + 2[HPO 4 2 - ] + 3[PO 4 3 - ] correct Explanation: To balance charge (iF the initial net charge is zero), simply add up all the species which have positive charge, and equate them to all the species which have negative charge. IF the magnitude oF charge is greater than 1, you must include a coe±cient to compensate. ²or example, [HPO 4 2 - ] should be 2[HPO 4 2 - ] in the charge balance equation. 003 6.0 points What is the [OH - ] oF 8 × 10 - 9 M NaOH? 1. 2 . 0016 × 10 - 7 M 2. 9 . 608 × 10 - 8 M 3. 8 × 10 - 9 M 4. 1 . 15096 × 10 - 7 M 5. 1 . 0408 × 10 - 7 M correct Explanation: NaOH is a strong base, so it completely dissociates. This adds OH - ions to solution, which initially will violate the equilibrium ex- pression For the dissociation oF water. Then some oF the OH - will be neutralized with the H + until the expression is obeyed again.
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Version 139 – Make up 2 – laude – (51635) 2 K w = [H + ][OH - ] 10 - 14 = (10 - 7 M - x ) · (10 - 7 M + 8 × 10 - 9 M - x ) = ([OH - ] - 8 × 10 - 9 M)[OH - ] [OH - ] = 1 . 0408 × 10 - 7 M 004 6.0 points What is K sp for HgI 2 , if its molar solubility is 2 . 2 × 10 - 10 mol / L? 1.
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Make up 2-solutions - Version 139 Make up 2 laude (51635)...

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