A student weighs 0.683 g of potassium acid phthalate, HKC_8H_4O_4 abbreviated KH

Question

A student weighs 0.683 g of potassium acid phthalate, HKC_8H_4O_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_2H_3O_2) are in the sample of vinegar? d. What is the percent acetic in the sample of vinegar?

Explanation / Answer

Q1.

a)

mol = mass/MW = 0.683 / 204.2 = 0.00334476 mol of KHP

b)

moles of NaOH used

1 mol of NaOH = 1 mol of KHP

0.00334476 mol of KHP = 0.00334476 mol of NaOH

c)

M = mol/V

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

now

mol = 0.00334476

so

M = (0.00334476 )/ (33.74 *10^-3) = 0.099133 M

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