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 )

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

h let me think