500 ml of NaOH analyte was neutralized with 10 ml of 0.5 M H_2, SO_4 What was th

Question

500 ml of NaOH analyte was neutralized with 10 ml of 0.5 M H_2, SO_4 What was the concentration of the NaOH in the analyte? How many ml of 0.5 M HCl would be required to neutralize the analyte from 1? Iodine with in solution to give a dark blue color This can used as an indicator for the titration of acid (Vitamin C). As long as there is vitamin C in solution iodine is reduced to iodide and does not react with starch to produce the blue I_f + ascorbic acid rightarrow 2I + dehydroascorbic acid An ascorbic acid solution made by dissolving vitamin C tablets in 2.00 ml of water is titrated with 2.50ml of 0.05 M I solution a) What is the molar concentration of the vitamin C solution? b) How many grams of vitamin Care in the solution? c) How many 500g vitamin C tablets are were used to make the solution?

Explanation / Answer

1) NaOH Molarity =M1

H2SO4 Molarity =2 M2 = 1.0 M {since it is a dibasic acid}

2 NaOH + H2SO4 Na2SO4 + 2 H2O

V1 M1 = V2x 2 M2

500 ml x M1 = 10 ml x 1.0 M

M1 = 10 /500 = 0.02 M

2)

NaOH Molarity =M1 = 0.01 M , Volume V1 = 500 ml

HCl Molarity =M2 = 0.5 M V2 = ?

V1 M1 = V2 M2

500 ml x 0.01 M = V2 x 0.5 M

V2 = 5 /0.5 = 10 ml

3)

1 mole I2 reacts with 1 mole ascorbic acid.

So V1 M1 = V2 M2

200 ml x M1 = 2.50 ml x 0.05 M

M1 = (2.50 ml x 0.05 M) / 200 = 0.000625 moles/litre = 0.000625 M

a) Molar concentration of ascorbic acid = 0.000625 M

b)Molecular weight of ascorbic acid = 176 g/mol

Weight of vitamin C in given solution = (0.000625 x 176 x 200) /1000 = 0.022 g

c)

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