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welcome x Dotson x Norther x strong A x C calculate x Di lonization x Calculate x C Chegg st x CA1.46 LE x M Inbox h x C D sapling learning.com /ibiscm s/mod/ibis/v iew.php?id 3596987 4/18/2017 11:55 PM A 52.9/100 Gradebook Assignment Information Attempts Score tr Print Calculator Periodic Table Available From Not Set Question 4/18/2017 11:55 PM Due Date: Sapling Learning Map Points Possible Calculate the pH for each of the following cases in the titration of 35.0 mL of 0.180 M LiOH(aq), Grade Category: Homework with 0.180 M HCl(aq) Description: Note: Enter your answers with two decimal places. Homework (10 tries, no MC Policies (a) before addition of any HCI deduct You can check your answers. (b) after addition of 13.5 mL of HCI You can view solutions when you complete or give Number up on any question. (c) after addition of 20.5 mL of HCl You have ten attempts per question. There is no penalty for incorrect answers. Number (d) after the addition of 35.0 mL of HCI O eTextbook O Help With This Topic e) after the addition of 42.5 mL of HCI O Web Help & Videos Number (f after the addition of 50.0 mL of HCl OTechnical Support and Bug Reports O Previous Give up Vi solution check Answer ONext Exit 12 PM Search right here 4/18/2017

Explanation / Answer

the reaction:

               HCl + LiOH --->LiCl + H2O

The key is that there is a 1:1 molar ratio between HCl and LiOH.

moles LiOH ---> (0.180 mol/L) (0.035 L) = 0.0063 mol

Determination of pH of the solutions:

a) Iniitial point (0 ml of HCl);

We have LiOH Concentration = 0.180 M = [OH-]

pOH = -log[OH-] = -log [0.180] = 0.745
pH = 14 - pOH = 14 - 0.745 = 13.26

b) After addition of 13.5 mL of HCl;
no. moles of of LiOH = 0.0063 mol - (0.0135 x 0.180) mol = 0.002439
total volume = 35 mL + 13.5 mL = 48.5 mL = 0.0485 L

Molarity = no. moles/ volume in L = 0.002439 / 0.0485 L = 0.0503 M
pOH = -log(0.0503) = 1.30
pH = 14 - pOH = 14 - 1.3 = 12.70

c) 20.5 mL of HCl

no. moles of of LiOH = 0.0063 mol - (0.0205 x 0.180) mol = 0.00261
total volume = 35 mL + 20.5 mL = 55.5 mL = 0.0555 L

Molarity = no. moles/ volume in L = 0.00261 / 0.0555 L = 0.047 M
pOH = -log(0.047) = 1.33
pH = 14 - pOH = 14 - 1.33 = 12.67

d) 35 mL of HCl

no. moles of of LiOH = 0.0063 mol - (0.035 x 0.180) mol = 0
This is the nuetralization point,So, [H+] = [OH-] = 10-7.
pH = 7.00

e) 42.5 mL of HCl

After nuetralization we have excess HCl present in the solution.
Therefore;
no. moles of of HCl = (0.0425 x 0.180) mol - 0.0063 mol = 0.00135 mol of HCl
total volume = 35 mL + 42.5 mL = 77.5 mL = 0.0775 L

Molarity = no. moles/ volume in L = 0.00135 mol / 0.0775 L = 0.0174 M
pH = -log(0.0174) = 1.76


f) 50 mL of HCl

no. moles of of HCl = (0.05 x 0.180) mol - 0.0063 mol = 0.0027 mol of HCl
total volume = 35 mL + 50 mL = 85 mL = 0.085 L

Molarity = no. moles/ volume in L = 0.0027 mol / 0.085 L = 0.0318 M
pH = -log(0.0318) = 1.50

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