A student dissolved 0.1916g of an unknown diprotic acid in 100mL of distilled wa

Question

A student dissolved 0.1916g of an unknown diprotic acid in 100mL of distilled water. The acid was then titrated with 0.1025M NaOH solution. The second equivalence point showed the sharpest change in pH, and so it was used to determine the molar mass of the unknown acid. The volume of NaOH needed to reach the equivalence point was 27.5mL.

1) Calculate the number of moles of NaOH used in the titration to reach the second equivalence point.

2) Calculate the number of moles of diprotic acid, based on the fact that we are examining the second equivalence point.

3) Calculate the Molar Mass of the diprotic acid.

Explanation / Answer

m = 0.1915 g of H2A

V = 100 ml

M = 0.1025 NaOH

VT = 27.5 ml

1)

NaOH used in titration

Mol = M*V = 0.1025*27.5 = 2.819 mmol of NaOH or 0.00282 mol of NaOH

2)

calculate number of moles

H2A + 2NaOH --> 2H2O + Na2A

2 mol of base = 1 mol of acid

therefore

1/2 *0.00282 = 0.00141 mol of H2A present

3)

calculate MW of acid

MW = mass/mol

mol = M*V = 0.00141 * 0.1 = 1.41*10^-4

MW = 0.1916/(1.41*10^-4) = 1358.86 g/mol

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