Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

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 Hint

Explanation / 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

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote