Chemistry Lab Help

Question

student prepares a dilute solution of sodium hydroxide, NaOH starting with 6 M sodium hydroxide. She then titrates a 1.372 g sample of KHP with the dilute sodium hydroxide solution, NaOH (aq), to a phenolphthalein end point. the titration required 21.84 mL of sodium hydroxide, NaOH (aq), to a phenolphthalein end point. If The titration requires 21.84mL of sodium hydroxide, NaOH, calculate the mass percent of acetic in the vinegar using the molar concentration for acetic acid, 0.8452m and assuming the density of the solution is 1.01g/mL.

Answer 1

Assuming KHP is potassium hydrogen phosphate, a monoprotic acid, then...

KHP + NaOH ==> NaKP + H2O i.e. a 1:1 molar ratio of NaOH to KHP

1.372 g KHP x 1 mol KHP/204.22 g = 0.006718 moles KHP

moles NaOH used = 0.006718 b/c 1 mol KHP = 1 mol NaOH

molarity of NaOH = 0.006718 moles/0.02184 L = 0.3076 M NaOH

Now, your question becomes vague again. Where does the acetic acid come in? What volume of acetic acid was titrated with the NaOH, or what mass? We need more information about the experiment carried out on the acetic acid.