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.
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
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1 answer