# Draw the ph vs volume of NaOH

Draw the ph vs volume of NaOH for titrating HOOCCH2CH2OH withNaOH. What is the ph at the endpoint? What will be different whenHOOCCH2COOH is titrated with NaOH? Please explain what you aredoing. Thank you

Titration plot for monoprotic acid

lets say we are titrating 50 ml of 1 M HOOCCH2CH2OH with 1 MNaOH

initial pH

Acid dissociates in solution,

HOOCCH2CH2OH <==> H+ + -OOCCH2CH2OH

let x amount dissociated

Ka = 3.1 x 10^-5 = x^2/1

x = [H+] = 5.57 x 10^-3 M

pH = -log[H+] = 2.25

acid remaining in solution = (1 M x 50 ml – 1 M x 12.5 ml)/62.5ml = 0.6 M

acid salt formed = 12.5 mmol/62.5 ml = 0.2 M

pH = pKa + log(base/acid)

= 4.51 + log(0.2/0.6) = 4.03

half-equivalence point

pH = pKa = -log(Ka) = 4.51

acid remaining in solution = (1 M x 50 ml – 1 M x 37.5 ml)/87.5ml = 0.143 M

acid salt formed = 37.5 mmol/87.5 ml = 0.43 M

pH = pKa + log(base/acid)

= 4.51 + log(0.43/0.143) =5.0

equivalence point

conjugate acid formed hydrolyzes

[conjugate acid] = 1 M x 50 ml/100 ml = 0.5 M

let A- be the conjugate acid

A- + H2O <==> HA + OH-

let x amount reacted

Kb = 1 x 10^-14/3.1 x 10^-5 = x^2/0.5

x = [OH-] = 1.27 x 10^-5 M

[H+] =1 x 10^-14/1.27 x 10^-5 = 7.9 x 10^-10 M

pH = -log[H+] = 9.10

excess [OH-] = 1 M x 150 ml/200 ml = 0.75 M

pOH = -log[OH-] = 0.125

pH = 14 – pOH = 13.87

Plot shown below

When the diacid HOOCCH2COOH is titrated instead, 2 moles of NaOHis used to reach equivalence point as opposed to 1 mole as seen inthe previous case.

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