What volume of a 0.00945-M solution of potassium hydroxide would be required to titrate 50.00 mL of a sample of acid rain with a H2SO4 concentration of 1.23 * 10-4 M. H2SO4 (aq) + 2KOH (aq) ⟶K2SO4 (aq) + 2H2O (l)

Question

Chemistry

Emmett

12 July, 07:35

What volume of a 0.00945-M solution of potassium hydroxide would be required to titrate 50.00 mL of a sample of acid rain with a H2SO4 concentration of 1.23 * 10-4 M. H2SO4 (aq) + 2KOH (aq) ⟶K2SO4 (aq) + 2H2O (l)

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Answers (1)

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Scarlet · Answer 1

1.30mL

Explanation:

The equation for the reaction is given below:

H2SO4 + 2KOH - > K2SO4 + 2H2O

From the equation above we obtained the following:

Mole of acid (nA) = 1

Mole of base (nB) = 2

The following data were obtained from the question:

Mb = 0.00945M

Vb = ?

Va = 50mL

Ma = 1.23 * 10^-4M

Using MaVa / MbVb = nA/nB, we can calculate the volume of KOH as illustrated below:

MaVa / MbVb = nA/nB

(1.23 * 10^-4 x 50) / 0.00945xVb = 1/2

Cross multiply to express in linear form

1.23 * 10^-4 x 50 x 2 = 0.00945xVb

Divide both side by 0.00945

Vb = (1.23 * 10^-4 x 50 x 2) / 0.00945

Vb = 1.30mL