A 10.0% potassium hydroxide solution (density 1.092 g / ml) was carefully added to an 8.3% sulfuric

A 10.0% potassium hydroxide solution (density 1.092 g / ml) was carefully added to an 8.3% sulfuric acid solution with a volume of 45.0 ml (density 1.05 g / ml) until the acid was completely neutralized. Determine the volume of the poured alkali solution. How many grams of potassium sulfate should be added to the resulting solution so that the mass fraction of salt in it becomes 10.0%?

H2SO4 + 2KOH = K2SO4 + 2H2O

Find the mass of the sulfuric acid solution.

m = p * V = 45 * 1.05 = 47.25 g

Now we calculate the mass and the amount of moles of sulfuric acid in the solution.

m (H2SO4) = 47.25 * 0.083 = 3.92 g

n (H2SO4) = m / M = 3.92 / 98.01 = 0.04 mol

Find the number of moles and the mass of KOH to neutralize all the acid.

n (KOH) = 2 * n (H2SO4) = 2 * 0.04 = 0.08 mol

m (KOH) = n (KOH) * M = 0.08 * 56.1 = 4.5 g

Find the mass of a 10% potassium hydroxide solution.

mр (KOH) = 4.5 * 100/10 = 45 g

Knowing the density of the solution, we find its volume.

V (KOH) = 45 / 1.092 = 41.2 ml

Now we will find the mass of potassium sulfate, which is formed during the reaction.

m (K2SO4) = n (K2SO4) * M = 0.04 * 174.3 = 7.0 g

The volume of the resulting solution is V (KOH) + V (H2SO4) = 41.2 + 45.0 = 86.2 ml

Mass fraction should be 10%, which means:

0.1 = m / 86.2

m = 8.62 g K2SO4

The solution already contains 7.0 g of salt, which means you need to add another 8.62 – 7.0 = 1.62 g of K2SO4.