Problem 2.7 Cakuulate the pH walues and draw the titration curve for the or 500
ID: 884064 • Letter: P
Question
Problem 2.7 Cakuulate the pH walues and draw the titration curve for the or 500 mL of oo10 Mapetc acid pKa titration 4.76) wh o 010 M KOH Part A Calculate the of the solution anerarnLof the titrant have been added. newer using two deolmal Expres your a plaaes. My Answers Part B calculate the of the solution aner 250 mLof the strant have been added newer ucing two deolmal ExpreGG your a plaaes. My Anwert Part C calculate the pH of the solution aner 490 mL or the titrant have been added Expreco your anewer using two dealmal plaaes. My Ancwere Part D calculate the at the solution aner 500 mL or the titrant have been added. Exprece your answer ucing two deolmal plaaos. My Answers 3Ive Part E calculate the pHofthe soluton atter 510 mL of the titrant have been added. ExpreEE your al ucing two dealmal plaoeG My Answers Part F calculate the of the solution after 750 mL of the titrant have been added. uclng two dealmal polsoec. ExpresE your a My Answers Part G This question wil be shown ather you complete previous question(s)Explanation / Answer
PaRt A.
No of moles of Aceticacid = 500*0.01/1000 = 0.005 mole
before addition of KOH
pH of weak acid = 1/2(pka-log C)
pka of aceticacdi = 4.76
Concentration of acetic acid = 0.005 M
= 1/2(4.76-log 0.005) = 3.53
Part B.
No of moles of Aceticacid = 500*0.01/1000 = 0.005 mole
no of moes of KOH added = 250*0.01/1000 = 0.0025 Mole
below the Half equivalence point pH depends up on Buffer formula
pH = pKa + log (salt/acid)
salt = KoH added = 0.01*250/750 = 0.00333 M
acetic acid concentration = 0.001*500/750 = 0.0066 M
pH = 4.76 + log(0.0033/0.0066-0.0033)
pH = 4.76 (at half equivalence point pH = pKa)
Part C
No of moles of Aceticacid = 500*0.01/1000 = 0.005 mole
no of moes of KOH added = 490*0.01/1000 = 0.0049 Mole
Concentration of salt = 0.0049*1000/990 = 0.00495 M
above the Half equivalence point pH depends up on salthydrolysis formula
here salt is formed by weak acid,strong base.
pH = 7+1/(pka+logC)
= 7+1/2(4.76+log 0.00495)
= 8.22
Part D
No of moles of Aceticacid = 500*0.01/1000 = 0.005 mole
no of moes of KOH added = 500*0.01/1000 = 0.005 mole
Concentration of salt = 100*0.01/1000 = 0.001 M
up on salthydrolysis formula
here salt is formed by weak acid,strong base.
pH = 7+1/(pka+logC)
= 7+1/2(4.76+log 0.01)
= 8.38
part E
No of moles of Aceticacid = 500*0.01/1000 = 0.005 mole
no of moes of KOH added = 510*0.01/1000 = 0.0051 mole
Concentration of excess KOH added = (510-500)*0.01/1010 = 9.9*10^-5 M
pOH = -log(9.9*10^-5) = 4.0
pH = 14-4 = 10
PaRT F
No of moles of Aceticacid = 500*0.01/1000 = 0.005 mole
no of moes of KOH added = 750*0.01/1000 = 0.0075 mole
Concentration of excess KOH added = (750-500)*0.01/1250 = 0.002 M
pOH = -log(0.002) = 2.7
pH = 14-2.7 = 11.3
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