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? (Density of vinegar is 1.005 g/mL) _________________

Answer 1

KHC_8 H_4O_4(ag) + NAoh(AG) rightarrow Na K C_8 H_4 O_4 pg) + H_2O(l) Moles of KHP = mass of KHp/molar mass of KHp = 0.683g/204.2 g mol = 3.345 times 10^-8 mol For 1 mole KHP we require 1 mole NaOH Therefore, moles of NaOH used = 3.545 times 10^-3 moles Molarity of the base = moles of NaOH/volume in L Volume of NaOH = .33.86 - 0.12 = 3.7.74 ml = 33.74 times 10^-3 L Molarity = 3.345 times 10^-5 moles/33.74 times 15^3 L = 0.099 H From part 1c, molarity of bar = 0.099 M Moles of NaoH used = (13.20 - 1.48 L)/100 times 0.099 mol L = 4.13 times 10^-3 moles CH_3COOH (eq) + NaOH(ag) rightarrow CH_3 Coo(ag) + H_2O moles of CH_3COOH = moles of NaOH used = 4.13 times 10^-3 moles grams of CH_3COOH = (4.15 times 10^-3 moles) times 50.05 g mol = 0.248 g