fak chem hw 7 - To (st22362) Homework 7 Fakhreddine (52420)...

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To (st22362) – Homework 7 – Fakhreddine – (52420) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. 001 1.0 points We fnd that 61.1 mL o± an HCl solution reacts exactly with 74.2 mL o± 1.371 M KOH solution. Calculate the molarity o± the HCl solution. 1. 0.832 M 2. 0.667 M 3. 3.33 M 4. 1.17 M 5. 1.66 M correct Explanation: V 1 = 61 . 1 mL V 2 = 74 . 2 mL M 2 = 1 . 371 M The balanced equation ±or the reaction is HCl + KOH KCl + H 2 O Molarity is moles solute per liter o± solution. We know the volume o± the HCl solution. I± we could fnd the moles o± HCl in the solution we could calculate the molarity. We start by calculating the moles o± KOH present: ? mol KOH = 74 . 2 mL soln × 1 L soln 1000 mL soln × 1 . 371 mol KOH 1 L soln = 0 . 1017 mol KOH Using the mole-to-mole ratio ±rom the bal- anced chemical equation we calculate the moles o± HCl needed to react with this amount o± KOH: ? mol HCl = 0 . 1017 mol KOH × 1 mol HCl 1 mol KOH = 0 . 1017 mol HCl This is the number o± moles o± HCl needed to react with the KOH and there±ore the num- ber o± moles present in the 61.1 mL o± HCl solution. We can calculate the molarity o± the HCl solution by dividing the moles o± HCl by the volume o± the HCl solution: ? M HCl = 0 . 1017 mol HCl 0 . 0611 L soln = 1 . 66 M HCl 002 1.0 points 200 mL o± 1.5 M HCl is mixed with 150 mL o± 3.0 M NaOH. HCl + NaOH NaCl + H 2 O What is the molarity o± the resulting salt solution? 1. 0.86 M NaCl correct 2. 0.45 M NaCl 3. 1.5 M NaCl 4. 2.25 M NaCl 5. 2.0 M NaCl Explanation: V 1 = 200 mL M 1 = 1 . 5 M V 2 = 150 mL M 2 = 3 . 0 M Our frst step is to determine the number o± moles o± each reactant present: ? mol HCl = 200 mL soln × 1 L soln 1000 mL soln × 1 . 5 mol HCl 1 L soln = 0 . 3 mol HCl ? mol NaOH = 150 mL soln × 1 L soln 1000 mL soln × 3 . 0 mol NaOH 1 L soln = 0 . 45 mol NaOH From the balanced chemical equation, we know that one mole o± HCl is needed to re- act with each mole o± NaOH. We don’t have
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To (st22362) – Homework 7 – Fakhreddine – (52420) 2 enough HCl to react all the NaOH, so HCl is our limiting reactant and will determine how much NaCl is produced: ? mol NaCl = 0 . 3 mol HCl × 1 mol NaCl 1 mol HCl = 0 . 3 mol NaCl We are asked to fnd the molarity o± the salt solution. Molarity is moles solute per liter o± solution. The total volume o± the resulting solution will be the combined volume o± the two original solutions that were mixed: ? mL NaCl soln = 200 mL NaCl + 150 mL NaCl = 350 mL NaCl We fnd the molarity o± NaCl in the solution by dividing the moles o± NaCl by the volume o± the solution: Molarity o± NaCl = 0 . 3 mol NaCl 350 mL soln × 1000 mL soln 1 L soln = 0 . 86 M NaCl 003 1.0 points Calculate the pH o± a solution when 0.200 moles o± solid NaOH is added to 2.00 liter o± 0.250 M HCl. Assume no change in volume. 1.
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This note was uploaded on 04/16/2010 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas at Austin.

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fak chem hw 7 - To (st22362) Homework 7 Fakhreddine (52420)...

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