04/02/1 This is a Numeric Entry question,/It is worth 3 points/You have unlimite

Question

04/02/1 This is a Numeric Entry question,/It is worth 3 points/You have unlimited attempts/There is no attempt pe 12 Question (3 points) Q See page 798 A 250 ml sample of a 0.1400 M solution of aqueous trimethylamine is titrated with a 0.1750 M solution of HCL.Calculate t e pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pk, of (CH3)3N-4.19 at 25°C v 5th attempt Part 1 (1 point) See Periodic Table See Hint pH after 10.0 mL of acid have been added- 9.81 Part 2 (1 point) pH after 20.0 mL of acid have been added- Part 3 (1 point) pH after 30.0 mL of acid have been added 4th attempt

Explanation / Answer

Trimethyl amine = 25.0ml of 0.1400M

number of moles of trimethyl mine = 0.1400Mx0.0250L = 0.0035 moles

PKb = 4.19

a) aftera addition of 10.0ml of HCL

HCl = 10.0ml of 0.1750M

number of moles of HCl =0.1750Mx0.010L= 0.00175 moles

          (CH3)3N + HCl ----------------- (CH3)3NH+Cl-

0.0035             0.00175                    0

-0.00175        - 0.00175                     +0.00175

0.00175             0                             +0.00175

after addtion of HCl

reamining number of moles of Trimiethyl amine = 0.00175 moles

number of moles of salt = 0.00175 moles

PKb= 4.19

POH= PKb + log[salt/base]

POH= 4.19 + log{0.00175/0.00175)

POH= 4.19

POH+PH= 14

PH= 14-POH

PH= 14-4.19

PH =9.81

B) after addition o f 20.0 ml of HCl

numberr o fmoles of HCl= 0.1750Mx0.020L =0.0035 ,moles

number o fmoles of base is eqaul to number o fmoles of acid

so it is equivalent point .

at equivalent point

PH= 7- 1/2[Pkb + logC]

total volume = 25.0+20.0= 45.0ml = 0.045L

C= number o fmoles/volume

C= 0.0035/0.045L= 0.0778M

PH= 7 - 1/2[4.19+log(0.0778)]

PH= 5.46

C) after addition of 30.0ml of HCl

number of moles of HCl= 0.1750Mx0.030L= 0.00525 moles

number of moles of acid is greatertah base. so bufffer is not formed. so the natutre of the solution is acidic,

remaining number of mole of acid= 0.00525 - 0.0035 = 0.00175 moles

total volume = 25.0+30.0= 55.0ml= 0.055L

[H+] = 0.00175/0.55 = 0.0318

[H+] = 0.0318M

-log[H+]= -log(0.0318)

PH= 1.497

PH= 1.50

Related Questions

Navigate

• Browse (All)
• Browse
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.