Question

Imagine you have prepared a 25.0 mL sample of .100 M glycine hydrochloride solution to use in a titration with NaOH.
a) Using your pKa1 value (2.38) and the Henderson-Hasselbach equation, calculate the ratio of the concentrations of ([C₂H₅NO₂]/[C₂H₆NO₂⁺]) present at pH= 2.50 for glycine hydrochloride solution.
b) What volume of .100M NaOH would you have to add during the titration to reach pH = 2.50?

Answer 1

Solution :-

part a) Henderson equation
pH= pka + log ([C₂H₅NO₂]/[C₂H₆NO₂⁺])

2.50 = 2.38 + log ([C₂H₅NO₂]/[C₂H₆NO₂⁺])

2.50 – 2.38 = log ([C₂H₅NO₂]/[C₂H₆NO₂⁺])

0.12 = log ([C₂H₅NO₂]/[C₂H₆NO₂⁺])

Antilog 0.12 = ([C₂H₅NO₂]/[C₂H₆NO₂⁺])

1.32 = ([C₂H₅NO₂]/[C₂H₆NO₂⁺])

So the ratis of the ([C₂H₅NO₂]/[C₂H₆NO₂⁺]) present at pH 2.50 is 1.32

b) What volume of .100M NaOH would you have to add during the titration to reach pH = 2.50?

Solution :-

Lets calculate the moles of the glycine hydrochloride

Moles = molarity * volume

            = 0.100 mol per L * 0.025 L

           = 0.0025 mol

b) Now lets calculate the moles of NaOH needed to reach the pH 2.50

pH= pka + log ([C₂H₅NO₂]/[C₂H₆NO₂⁺])

2.50 = 2.38 + log ([x]/[0.0025])

2.50 – 2.38 = log ([x]/[0.0025])

0.12 = log ([x]/[0.0025])

Solving for x we get 0.002818 mol

Now lets calculate the volume of the NaOH

Volume of NaOH = moles / molarity

                                = 0.002818 mol / 0.100 L

                               = 0.02818 L

0.02818 L * 1000 ml / 1 L = 28.2 ml

So 28.2 ml NaOH is needed.