Question

After the first trial, in order to maximize the titration precision, the sample size of pure KHC₈H₄O₄ or unknown for a trial should be adjusted so 20-23 mL of NaOH are delivered in both part B and C. This can be done as noted on page 82 in your lab manual.
If 6.61 mL NaOH was used in the titration of a 0.663 g sample use ratios to determine mass of sample needed to use 20.50 mL of NaOH in the next trial.

Answer 1

The reaction between KHP and NaOH is shown by the balanced equation

KHC₈H₄O₄ + NaOH ===> NaKC₈H₄O₄ + H₂O.

KHP stands for potassium hydrogen phthalate, which has the chemical formula KHC8H4O4.

let us find the molarity of NaOH

0.663 g = 0.663 /204.22 = 0.0032464 Moles

Moles of NaOH = 0.0032464

Molality of the NaOH soluiton = 0.0032464 x 1000 /6.61 = 0.4911 M

Then calculate moles for 20.5 ml of 0.4911 M NaOH = 20.5 x 0.4911 = 0.01 Moles

Mass of KHC₈H₄O₄ for 0.01 Moles =   0.01 x 204.22 = 2.056 gm

Hence 2.056 gm of needs 20.5 ml of 0.4911 M NaOH

answered by: anonymous