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le Edt View History Bookmarks Tools Help a Masteringchemistry Assig 4 httpsllsession.masteringchemistry.com/myet/temviewlassignmentproblemlD: e q, seordh MW Chem 1412 Fall 2016 Assignmen133 chap17 IAranensedL Problem 17.45 Problem 17.45 36.0 ml, Express your answer using two decimal places. A 35.0-mL sample of 0.150 M acetic acid (CH COOH) is titrated with 0.150 M NaoH solution. Calculate the pH after the folowing volumes of base have been added. pH 11.5 Submit Answers Gheup Incorrect. Try Again: 9 attempts remaining Part E 355ml. Express your answer using two decimal places. Sort By Answers Groove Folder synchronization Part F Graphics Properties. Express your answer using two decimal places. Graphics options

Explanation / Answer

Pka of acetic acid = 4.76, Ka = 1.74*10-5

Moles of CH3COOH in 0.15M of 35ml = molarity* Volume (L)= 0.15*35/1000=0.00525

The reaction is CH3COOH+ NaOH-----à CH3COONa +H2O

1 mole of acetic acid required 1 moles of sodium hydroxide to form 1 mole of sodium acetate

Moles of NaOH in 35 ml of 0.15M= 0.15*35/1000 =0.00525

Moles of sodium acetate formed = 0.00525 volume after mixing = 35+35=70ml =0.07 L

Concentration of sodium acetate = 0.00525/0.07=0.075

The reaction is CH3COONa+ H2O--à CH3COO- + OH-

Kb= [OH-] [CH3COO-]/[CH3COONa]

Kb= 10-14/ Ka =10-14/1.74*10-5 = 5.75*10-10

let x= drop in concentration of CH3COONa to reach equilibrium

  5.75*10-10 = x2/ (0.075-x)

When solved, x =2.07/105 = [OH-]

pOH= 4.7, pH= 14-4.7= 9.3

2.

moles of   NaOH in 35.5ml, 35.5*0.15/1000 =0.005325

excess is NaOH and excee moles of NaOH=0.005325-0.00525=7.5*10-5 Volume after mixing = 35+35.5= 71.5ml =0.0715L, Cocnentration = 7.5*10-5/0.0715=0.001049

pOH= 2.97, PH= 14-2.97= 11.03   (NaOHionized completely)

3. moles of   NaOH in 50l, 50*0.15/1000 =0.0075

excess is NaOH and excee moles of NaOH=0.0075-0.00525=0.00225Volume after mixing = 35+60= 95ml =0.095L, Cocnentration = 0.00225/0.095=0.0237

pOH=1.53 PH= 14-1.63= 12.37

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