3/14/2018 09:01 AM A 81/100 3/9/2018 04:44 PM Gradebook Calculator-4 Periodic Ta
ID: 1026582 • Letter: 3
Question
3/14/2018 09:01 AM A 81/100 3/9/2018 04:44 PM Gradebook Calculator-4 Periodic Table Question 20 of 28 Map Sapling Learning macmillan learning Calculate the pH for each of the following cases in the titration of 35.0 mL of 0.110 M LiOH(aq) with 0.110 M HI(aq) Number Note: Enter your answers with two decimal places (a) before addition of any HI Number (b) after addition of 13.5 mL of Hl Number (c) after addition of 23.5 mL of Hl Number (d) after the addition of 35.0 mL of HI Number (e) after the addition of 44.5 mL of HI Number (f) after the addition of 50.0 mL of HI Previous O Check AnswerNext Exit Y HintExplanation / Answer
answer:
a) Poh = -log[OH]
[OH-] = 0.11
Poh = -log[0.11] = 0.9586
Ph = 14- Poh = 14 - 0.9586 = 13.041
b) M =( M1V1 - M2V2)/(V1+V2)
M = (35*0.11 - 13.5*0.11)/(35+13.5)
M = 0.049
no. of moles of LiOH is more so the resultant solution is basic in nature
[OH-] = 0.049
Ph = 14-{-log[OH-]}
Ph = 14 - {-log(0.049)}
Ph = 12.69
.........................................................................................
C) M =( M1V1 - M2V2)/(V1+V2)
M = (35*0.11 - 23.5*0.11)/(35+23.5)
M = 0.022
no. of moles of LiOH is more so the resultant solution is basic in nature
[OH-] = 0.022
Ph = 14-{-log[OH-]}
Ph = 14 - {-log(0.022)}
Ph = 12.34
..................................................................................................
d) M =( M1V1 - M2V2)/(V1+V2)
M = (35*0.11 - 35*0.11)/(35+13.5)
no. of moles of LiOH i= no.of moles of HCl
that means the LiOH is neutralised by HCl
so Ph of the solution is 7
.......................................................................................
e) M =( M1V1 - M2V2)/(V1+V2)
M = (44.5*0.11 - 35*0.11)/(35+44.5)
M = 0.0131
no. of moles of HCl is more than the no.of moles of LIOH so the resultantant solution is acidic in nature
[H+] = 0.0131
Ph = -log[H+]
Ph = -log(0.0131)
Ph = 1.88
f)
M =( M1V1 - M2V2)/(V1+V2)
M = (50*0.11 - 35*0.11)/(35+50)
M = 0.0194
no. of moles of HCl is more than the no.of moles of LIOH so the resultantant solution is acidic in nature
[H+] = 0.0194
Ph = -log[0.0194]
Ph = 1.712
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