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fak chem hw 7

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 – find all choices before answering. 001 1.0 points We find that 61.1 mL of an HCl solution reacts exactly with 74.2 mL of 1.371 M KOH solution. Calculate the molarity of 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 for the reaction is HCl + KOH KCl + H 2 O Molarity is moles solute per liter of solution. We know the volume of the HCl solution. If we could find the moles of HCl in the solution we could calculate the molarity. We start by calculating the moles of 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 from the bal- anced chemical equation we calculate the moles of HCl needed to react with this amount of KOH: ? mol HCl = 0 . 1017 mol KOH × 1 mol HCl 1 mol KOH = 0 . 1017 mol HCl This is the number of moles of HCl needed to react with the KOH and therefore the num- ber of moles present in the 61.1 mL of HCl solution. We can calculate the molarity of the HCl solution by dividing the moles of HCl by the volume of the HCl solution: ? M HCl = 0 . 1017 mol HCl 0 . 0611 L soln = 1 . 66 M HCl 002 1.0 points 200 mL of 1.5 M HCl is mixed with 150 mL of 3.0 M NaOH. HCl + NaOH NaCl + H 2 O What is the molarity of 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 first step is to determine the number of moles of 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 of HCl is needed to re- act with each mole of 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 find the molarity of the salt solution. Molarity is moles solute per liter of solution. The total volume of the resulting solution will be the combined volume of the two original solutions that were mixed: ? mL NaCl soln = 200 mL NaCl + 150 mL NaCl = 350 mL NaCl We find the molarity of NaCl in the solution by dividing the moles of NaCl by the volume of the solution: Molarity of 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 of a solution when 0.200 moles of solid NaOH is added to 2.00 liter of 0.250 M HCl. Assume no change in volume. 1. 1.40 2. 0.82 correct 3. 0.59 4. 1.25 5. 1.53 Explanation: n NaOH = 0 . 200 mol n HCl = (2 . 0 L)(0 . 250 M) = 0 . 5 mol NaOH + HCl Na + + Cl - +H 2 O ini 0.200 0 . 5 0 0 Δ 0 . 200 0 . 200 0.200 0.200 fin 0 0 . 3 0.200 0.200 HCl is a strong acid, and Na + and Cl - are spectator ions.
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