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i Chrome File Edit View History Bookmarks People Window Help 53%D Fri 2:33 PM QOE Kean University-CHEM 1084xwww C www.sapli A Solutión ibiscms/mod/ibis/view.php?id 4322142 Sapling Learning Jump to... (2) Christina Brenckman , macmillan learning Sapling Learning Kean University - CHEM1084- Fall17-RUBINSTEIN Activities and Due DatesRequired HW 22 My Assignment Resources D 11/18/2017 10:30 AM 97.2/1011/16/201 7 06:53 PM # Attempts Score Gradebook Assignnent Information PrintCalauistor Periodic Table Question 7 of 10 Available From 11/10/2017 01:00 PM Due Date Points Possible: 10 Grade Category: Required HW 100 Sapling Learning Map 11/18/2017 10:30 AM What is the nickel ion concentration in a solution prepared by mixing 411 mL of 0,415 M nickel nitrate with 465 mL of 0.293 M sodium hydroxide? The K of nickel hydroxide is 1 x10-15 Number Policies: 04 You can check your answers. You can view solutions when you complete or give up on any question. You can keep trying to answer each question until you get it right or give up You lose 5% of the points available to each answer in your question for each incorrect attempt at that answer 100 10 1 O eTextbook Help With This Topi There is a hint available! View the hint by clicking an the O Web Help & Videos bottom divider bar Cl -on the divider bar again to hide the hint. Glase Technical Support and Bug Reports OPrenous Gve Up & View Solton Check Answer ) Next Ex- © 2011-2017 Sapling Learning, Inc. about uscareersprivacy policy terms of usecontactus help

Explanation / Answer

Reaction goes as follows

Ni(NO3)2 + 2 NaOH === Ni(OH)2 +2 NaNO3

let´s calculate the moles of each reactant with moles = Molarity * volume

moles of Ni(NO3)2 = 0.411 * 0.415 = 0.170565 available moles

moles of NaOH = 0.465 * 0.293 = 0.136245 available moles

as we can see we have more moles of nitrate than NaOH and we needed twice ammount of NaOH

some of the nitrate will remain untouched, to find the ammount of Ni(NO3)2 look at the stoichiometry

1 mole of nitrate needs 2 moles of NaOH

7 moles of nitrate will need 14 moles of NaOH

0.07 moles of nitrate will need 0.14 moles of NaOH so

we have 0.136245 moles of NaOH

moles of nitrate reacting is = 0.136245 / 2 = 0.0681225 moles

moles of nitrate remaining = 0.170565 - 0.0681225 = 0.1024425 moles

now look at the dissociation reaction

Ni(NO3)2 ===== Ni+2 + 2 NO3-1

0.1024425 moles of Ni(NO3)2 will produce 0.1024425 moles of Ni+2

now we need to get the total volume = 411 + 465 = 876 ml , divide it by 1000 to get 0.876 L

Molarity = moles / Volume

Molarity = 0.1024425 / 0.876 = 0.1169 M, this will be the concentration of Ni

you don´t need to do ksp calculations because the ammount of Ni coming from the precipitate won´t make a significant difference since you have 1 x 10-15 you will get a s value with an order of 10-5, it won´t make any difference against a value of 0.11M

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