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

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.

2b2t

explanation:

2b2t