You titrate a sample of the acid containing 0.367 g of the acid using 0.3341 M N

Question

You titrate a sample of the acid containing 0.367 g of the acid using 0.3341 M NAOH solution. Your initial buret reading is 1.06 mL and your final buret reading is 25.83 mL. Determine possible molar mass values for the monoprotic and diprotic cases

MM HA: ___________

MM H2A: ___________

Since you has no real information about the unknown acid, you worry that you may not have chosen an appropriate indicator for the titration. Assume that you have a 10% uncertainty in the molar mass values. Calculate the possible range of molar mass values for each case

Explanation / Answer

HA + NaOH ------> Na+ + A- + H2O

1 mole of HA reacts with 1 mole of NaOH

M1V1/n1 = M2V2/n2

M1 = molarity of HA

V1 = volume of acid solution = 20 mL since it is atitration.

n1 = number of moles of HA = 1

M2 = molarity of NaOH

V2 = volume of NaOH solution usedup in the reaction

n2 = number of moles of NaOH = 1

M1 * 20/1 = 0.3341 * (25.83-1.06)/1

M1 = molarity of HA = 0.4138

molarity of acid = mass of the acid/molar mass of the acid *1000/volume of acid solution

M1 = molarity of acid = 0.367/MW * 1000/20

0.4318 = 0.367/MW * 1000/20

MW = molar mass of the compound = 42.4965 g/mol

If 10% of uncertainity is ther then the error in molar mass is 10/100 = 0.1 amount

Molecular weight = 42.4965 +/- 0.1 g/mol

H2A + 2NaOH ------> 2Na+ + A- + 2H2O

1 mole of HA reacts with 2 mole of NaOH

M1V1/n1 = M2V2/n2

M1 = molarity of HA

V1 = volume of acid solution = 20 mL since it is atitration.

n1 = number of moles of HA = 1

M2 = molarity of NaOH

V2 = volume of NaOH solution usedup in the reaction

n2 = number of moles of NaOH = 2

M1 * 20/1 = 0.3341 * (25.83-1.06)/2

M1 = molarity of H2A = 0.2069

molarity of acid = mass of the acid/molar mass of the acid *1000/volume of acid solution

M1 = molarity of acid = 0.367/MW * 1000/20

0.2069 = 0.367/MW * 1000/20

MW = molar mass of the compound = 88.69 g/mol

If 10% of uncertainity is ther then the error in molar mass is 10/100 = 0.1 amount

Molecular weight = 88.69 +/- 0.1 g/mol

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