The titration of 25.0 mL of an unknown concentration H2SO4 solution requires 83.6 mL of 0.12 M LiOH solution for complete neutralization. What is molarity (M) of the H2SO4 solution

0.20M

Explanation:

Step 1:

Data obtained from the question. This include the following:

Volume of acid, H2SO4 (Va) = 25mL

Molarity of acid, H2SO4 (Ma) =.?

Volume of base, LiOH (Vb) = 83.6mL

Molarity of base, LiOH (Mb) = 0.12 M

Step 2:

The balanced equation for the reaction. This is given below:

H2SO4 + 2LiOH —> Li2SO4 + 2H2O

From the balanced equation above,

Mole ratio of the acid, H2SO4 (nA) = 1

Mole ratio of the base, LiOH (nB)= 2

Step 3:

Determination of the molarity of the acid, H2SO4. This can be obtain as follow:

MaVa/MbVb = nA/nB

Ma x 25 / 0.12 x 83.6 = 1/2

Cross multiply

Ma x 25 x 2 = 0.12 x 83.6

Divide both side by 25 x 2

Ma = (0.12 x 83.6) / (25 x 2)

Ma = 0.20M

Therefore, the molarity of the acid, H2SO4 is 0.20M.

the answer would be b i believe

that's what i got you can count the number of atoms