A student weights 0.683 g of potassium acid phthalate, HKC_8 H_4 O_4, abbreviate

Question

A student weights 0.683 g of potassium acid phthalate, HKC_8 H_4 O_4, abbreviated KHP (MM = 204 2 g/mol) and titrates it with NaOH The initial buret reading is 0.12 ml and the final buret reading at the end-point is 33.86 mL a. How many moles of KHP were used? b. How many moles of NaOH were used? c. What is the molarity of the base? The student then titrates 5.00 mL of vinegar with the same base. The initial buret reading is 1 48 mL and the final buret reading at the end-point is 43.20 mL. a. How many moles of base were used? b. How many moles of acid were used? c. How many grams of acetic acid (HC_2 H_3 O_2) are in the sample of vinegar? d. What is the percent acetic in the sample of vinegar?

Explanation / Answer

Q1.

a)

moles of KHP

mol = mass/MW = 0.683/204.2 = 0.0033447 mol of KHP

b)

moles of NaOH

ratio is 1:1 so

0.0033447 mol of KHP = 0.0033447 mol of NaOH

c)

V =Vf-Vi = 33.86-0.12 = 33.74 mL

M = mol/V = 0.0033447 / (33.74 *10^-3)

M = 0.09913159 mol per liter

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