# 250 ml of a 5 molar and 100 ml of a 0.8 molar sulfuric acid solution are mixed. The resulting volume was

**250 ml of a 5 molar and 100 ml of a 0.8 molar sulfuric acid solution are mixed. The resulting volume was added up to 3 liters. Find the molar concentration of the equivalent of the resulting solution**

Finding the molar concentration of the equivalent of the resulting solution is carried out according to the scheme:

M (H2SO4) = 98 g / mol;

feq (H2SO4) = 1/2 = 0.5;

ME (H2SO4) = fEq (H2SO4). M (H2SO4) = 0.5. 98 = 49 g / mol.

1) Calculate the total amount of sulfuric acid when mixing solutions, we get:

n (H2SO4) = (5. 0.25) + (0.8. 0.1) = 1.33 mol.

From here

2) Find the mass of sulfuric acid:

m (H2SO4) = n (H2SO4). M (H2SO4) = (1.33. 98) = 130.34 g.

3) Determine the molar concentration of the equivalent of sulfuric acid in a 3-liter solution of it according to the formula:

CH (H2SO4) = mH2SO4) / [ME (H2SO4). V] where

CH (H2SO4) – molar concentration of the equivalent of a solution of sulfuric acid (mol / l); V is the volume of the total acid solution.

CH (H2SO4) = 130.34 / (49.3) = 0.8866 mol / L (mol · L-1).

Answer: CH (H2SO4) = 0.8866 mol / l