Question

Orange and grapefruits are known as citrus fruits because their acidity comes mainly from citric acid, H3C6O7. Calculate the concentration of citric acid in a solution if a 30.0 Ml sample is completely neutralized by 15.10 Ml of 0.0100 M KOH. Assume all three acid hydrogens are neutralized in the reaction.

Moles of KOH reacted ?

Moles of citric acid reacted ?

Concentration of citric acid ?

What is the oxidation state of C in citric acid H3C6O7?

Answer 1

Given data: At neutralization point,

For KOH , M1 =0.01 M and V1 = 15.01 mL = 15.01 x 10^-3 L …(since, 1mL = 10^-3 L)

For Citric acid, M2 =?    If    V2 = 30.0 mL = 30.0 x 10^-3 L

We know that at point of neutralization Milimoles of base = Milimoles of acid.

i.e. M1V1 =M2V2

This is called as molar equation

We write therefore,

Milimoles of base = M1V1 = 0.01 x 15.01 = 0.1501 Milimoles

Moles of base = 0.01 x 15.01 x 10^-3 L = 0.1501 x 10^-3   = 1.501 x 10^-2 moles KOH. Base reacted.

At neutralization same 1.501 x 10^-2 moles of citric acid reacted.

Now,

M2V2 = M1V1.

M2 x 30.0 = 0.1501

M2 = 0.1501 / 30.0

M2 = 5.003 x 10^-3 M.

M2 = 5.003 x 10^-3

Concentration of citric acid is 5.003 x 10^-3 M.

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Calculation of O.S. of C in H3C6O7

We have net charge on H3C6O7 = 0

O.N. of H = +1

O.N. of O = -2

O.N. of C = p

We write therefor,

[3 x (+1) ] + [6 x (p)] +[7 x (-2)] = 0

(3 + 6p -14) = 0

6p -11 = 0

6p = +11

P = +11/6

P = +1.833

Hence Oxidation state of C in H3C6O7 is +1.833

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