Question

In a titration, 25.0 of dilute sulfuric acid was used to react completely with 20.0 of 0.400 mol/ aqueous sodium hydroxide.

2 NaOH (aq) + (aq) → (aq) + 2 (1) (a)Calculate the number of moles of sodium hydroxide that reacted.

(b) Calculate the number of moles of sulfuric acid that reacted with the sodium hydroxide. (c) What is the concentration of sulfuric acid in g/

Answer 1

Step-by-step explanation:

The balanced chemical equation for the reaction between sulfuric acid and sodium hydroxide is:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

From the equation, we can see that one mole of sulfuric acid reacts with two moles of sodium hydroxide. Therefore, the number of moles of sodium hydroxide used in the reaction is:

n(NaOH) = c(NaOH) x V(NaOH) = 0.400 mol/L x 0.0200 L = 0.00800 mol

Since one mole of sulfuric acid reacts with two moles of sodium hydroxide, the number of moles of sulfuric acid used in the reaction is twice that of sodium hydroxide:

n(H2SO4) = 2 x n(NaOH) = 2 x 0.00800 mol = 0.0160 mol

The concentration of the sulfuric acid can be calculated by dividing the number of moles by the volume used in the titration:

c(H2SO4) = n(H2SO4) / V(H2SO4) = 0.0160 mol / 0.0250 L = 0.640 M

Therefore, the concentration of the dilute sulfuric acid is 0.640 M.