In a titration if 25.0cm3 of a 0.01moldm3 sodium hydroxide solution required 10.0cm3 of a tetraoxosulphate(vi)acid solution for complete neutralization determine the moles of sodium hydroxide used in the reaction
To determine the moles of sodium hydroxide used in the reaction, we can use the equation:
NaOH + H2SO4 -> Na2SO4 + H2O
From the equation, we know that 1 mole of sodium hydroxide reacts with 1 mole of tetraoxosulphate(VI) acid.
Given:
Volume of sodium hydroxide solution = 25.0 cm^3
Molarity of sodium hydroxide solution = 0.01 moldm^-3
Volume of tetraoxosulphate(VI) acid solution = 10.0 cm^3
First, let's calculate the moles of tetraoxosulphate(VI) acid used in the reaction:
Moles of H2SO4 = Molarity × Volume
= 0.01 moldm^-3 × 10.0 cm^3
= 0.1 mol
Since the stoichiometric ratio between sodium hydroxide and tetraoxosulphate(VI) acid is 1:1, the moles of sodium hydroxide used in the reaction will be the same as the moles of tetraoxosulphate(VI) acid:
Moles of NaOH = 0.1 mol
Therefore, 0.1 moles of sodium hydroxide were used in the reaction.