The weak base ethanolamine, HOCH2Ch2NH2 can be titrated with HCl. HOCH2CH2NH2(aq

Question

The weak base ethanolamine, HOCH2Ch2NH2 can be titrated with HCl. HOCH2CH2NH2(aq) + H3O+---->HOCH2CH2NH3+(aq) + H2O(l) Assume you have 25.0mL of a 0.010M solution of ethanolamine and titrate it with 0.0095M HCl. (Kb for ethanolamine is 3.2X10^-5) a) What is the pH of the ethanolmine solution before the titration begins? b)What is the pH at the equivalence point? c) What is the pH at the halfway point of the titration? d) Calculate the pH of the solution after adding 5.00,10.0, and 20.0, and 30.0mL of the acid. Thanks!

Explanation / Answer

a)for weak base

pOH=0.5(pkb-log(c))=0.5*(4.494-log(0.01))=3.247

pH=14-pOH=10.75

b)

vlume of HCl neede for titration=26.315mL

at equivalence point only HOCH2CH2NH3+ will be there

ka for this=Kw/Kb=3.125*10^(-10)

c for this acid=0.01*25(25+26.31)=4.8723*10^(-3)

so pH=0.5(pKa-log(c))=5.9086

c)

at halfway point of the titration

solution will be a buffer

also [ethanolamine]=[HOCH2CH2NH3+]

so from handerson hasselbach equation

pOH=pKb + log(BH+/B)

pOH=pKb=4.4948

pH=9.5052

d)

5mL

moles of ethanolamine=(0.01*0.025 - 0.005*0.0095)=2.025*10^(-4)

mole of conjugate acid=(0.0095*0.005)=4.75*10^(-5)

so pOH=pKb+log(conjugate acid/ethanolamine)

pOH=3.865

pH=10.134

10mL

pOH=4.4948+log(0.0095*0.01/(0.01*0.025 - 0.0095*0.01)=4.2821

pH=9.7178

20mL

pOH=4.4948 +log(0.0095*0.02/(0.01*0.025-0.0095*0.02)=4.995

pH=9.0045

30ml

only Hcl and conjugate acid will be there.

moles of HCl=(30-26.315)*10^(-3)*0.0095=3.5*10^(-5)

c=6.365*10^(-4)

pH=-log(c)=3.1962

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