Potentiometric Titrations - Post Lab Calculations Data: - Mass of unknown: 0.250

ID: 1040973 • Letter: P

Question

Potentiometric Titrations - Post Lab Calculations

Data:

- Mass of unknown: 0.2500g

- Concentration of NaOH: 0.1000M

- Volume of NaOH at equivalnce point: 12.23mL

- Moles of NaOH at equivalnce point: 0.1000 moles x 0.01223L = 1.223x10^-3mL

- Moles of unknown acid at equivalnce point: ?

- Molecular Weight of unkown: ?

- Volume of NaOH added at half-equivalence: 12.23/2=6.115mL

Kindly help solve the below, showing all calculation steps:

1. Moles of unknown at equivalence point (show calculation)

2. Molecular weight of unknown acid (show calculation)

3. pH at half-equivalance point (show calculation)

4. Ratio of [B]/[Hb] at half-equivalence point (show calculation)

5. pKa (show calculation)

6. Ka (show calculation)

Thank you in advance!

1. Moles of unknown acid at equivalence point=mol of NaOH at equivalence point (for monoprototic acid,mol NaOH/mol acid=1:1)

So, Moles of unknown acid at equivalence point= 0.1000M*0.01223L=0.001223 mol

2. Molecular weight of unknown acid =M=mass acid/mol acid=0.2500g/(0.001223 mol)=204.415 g/mol

3. pH at half-equivalance point =pka +log [base]/[acid] [henderson-hasselbach eqn]

At half -eqv ,[base]=[acid]

pH=pka

4. Ratio of [B]/[HB] at half-equivalence point

pH =pka +log [base]/[acid]

But at half -eqv, pH=pka

So,log [base]/[acid] =log [B]/[HB]=0

or, [B]/[HB]=10^0=1

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