A monoproptic weak acid was titrated against kaOH. The concentration of the weak

Question

A monoproptic weak acid was titrated against kaOH. The concentration of the weak acid solution was 6.01 . The volume of acid used was 25 mL. The NaCH solution had a molarity of 0.125 M. The titration gave the following results. Volume of NaOH used to reach equivalence point = 10.8 mL pH at equivalence point = 0.76 pH at half equivalence point = 4.65 Determine pKa for the weak acid How many moles of the weak acid is there is 25 mL of solution to be ? Calculate the molar acid of the weak acid A digrotic weak acid was titrated against NaOH. The concentration of the weak acid solution was 10.36 g/L. The volume of acid used was 25.0 mL. The NaOH solution had a molarity of 0.125 M. The volume of NaOH is 25.0 ML at the second equivalence point, while pH is 11.62. The pH at the first equivalence point is 6.97. The pH at half first equivalence point is 3.78. Determine pKa1 and pKa2 of the weak acid Calculate the molar mass of the weak acid

Explanation / Answer

1. Monoprotic acid

1) pKa of acid = pH at first half equivalence point = 4.65

2) moles of HA in 25 ml = 0.125 M x 10.8 ml = 1.35 mmol

3) molar mass of acid = g/moles = 6.01 x 1000/1.35 x 40 = 111.30 g/mol

2. Diprotic H2A acid

1) pKa1 = pH at first 1/2 euivalence point = 3.78

pKa2 = 11.34

2) moles of acid = 0.125 M x 25 ml/2 = 1.5625 mmol

molar mass of acid = 10.36 x 1000/40 x 1.5625 = 165.76 g/mol

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