Question

A 35.0-ml sample of 0 150 Macetic acid (CH,COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added. Part A O mL Express your answer using two decimal places. KALOOB ? pH Submit Request Answer Part B 17.5 ml. Express your answer using two decimal places EVO AZOO pl Subrni an

Answer 1

part A

HA= CH₃COOH

A^- = CH₃COO^-

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Het = [ex] = CXNka t = vac -log H* = -12 log ka - lage / pH pika - Noge. DH 2 Scanned with CamScanner
FROM LITERATURE</sup>

^K_a =1.8*10^-5

^{Concentration C=0.150 M}

pK_a =-log(K_a) =-log(1.8*10^-5)

pK_a =4.74

pH=1/2(4.74-0.150)

pH=4.59/2

pH= 2.295

part B

CH₃COOH+NaOH$\rightarrow$CH₃COONa +H₂O

calculate the number of moles of acid ,base,salt

for acid

by defination of molarity

1000 ml of 0.150 M of acetic acid contains =0.150 moles of acetic acid

1 ml of 0.150 M of acetic acid contains =0.150/1000 moles of acetic acid

35 ml of 0.150 M of acetic acid contains =0.150/1000*35 moles of acetic acid

35 ml of 0.150 M of acetic acid contains =5.25 milli moles of acetic acid

for Base NaOH

1000 ml of 0.150 M of NaOH contains =0.150 moles of NaOH

1 ml of 0.150 M of NaOH contains =0.150/1000 moles of NaOH

17.5 ml of 0.150 M of NaOH contains =0.150/1000*17.5 moles of NaOH

17.5 ml of 0.150 M of NaOH contains =2.625 milli moles of NaOH

from reaction we see

1mole of acid react with 1 mole of base to give 1 mole of salt

since moles of base is less

moles of CH₃COONa=moles of NaOH=2.625 mmol

moles of acid left =moles of acid initially-moles of acid reacted

moles of acid left =(5.25-2.625) mmol=2.625mmol

using henderson equation

volume V=V_acid +V_NaOH =(35+17.5)ml =52.5 ml

concentration =moles/volume

[salt]=moles of CH₃COONa/volume of solution

[salt]=2.625mmol/52.5ml

[acid]=moles of CH₃COOH/volume of solution

[acid]=2.625mmol/52.5ml

pH=pK_a +log [salt]/[acid]

pH=4.74+log{(2.625mmol/52.5ml)/(2.625mmol/52.5ml)}

pH=4.74+ log 1

pH=4.74+ 0

pH=4.74