Question

Determine the %KHP in an unknown sample (mass = 645mg) that required an average of 15.27 mL of 0.0988 M NaOH to reach the end point. Show all calculations.

Answer 1

Here, 15.27 mL = 0.01527 L , 645 mg = 0.645 g , 0.0988 M = 0.0988 mol /L

moles of NaOH used = moles of KHP(Potassium hydrogen phthalate)

moles of NaOH used = (Volume of NaOH used) x (Concentration of NaOH)

Now,

moles of NaOH used = (0.01527 L) (0.0988 mol/L) = 0.00151moles

moles of NaOH used = moles of KHP = 0.00151 moles

Again,

mass of KHP in unknown = moles of KHP x formula weight of KHP(Potassium hydrogen phthalate)
= (0.00151 moles) x (204.22 g/mol) = 0.3084 g KHP (Potassium hydrogen phthalate)

Finally,

% KHP(Potassium hydrogen phthalate) in unknown sample = (mass KHP in unknown)/weight of unknown) x100%

= (0.3084 g /0.645 g) x 100%

= 0.4781 x 100 %

= 47.81 %