What volume of 0.50m of H2SO4 is required to neutralise completely 2.80g of KOH
To determine the volume of 0.50 M H2SO4 required to neutralize 2.80 g of KOH, we need to use the balanced chemical equation for the reaction between H2SO4 and KOH:
H2SO4 + 2 KOH -> K2SO4 + 2 H2O
From the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of KOH.
First, we need to determine the number of moles of KOH using its molar mass:
Molar mass of KOH = 39.10 g/mol (atomic mass of K) + 16.00 g/mol (atomic mass of O) + 1.01 g/mol (atomic mass of H)
= 56.11 g/mol
Number of moles of KOH = mass of KOH / molar mass of KOH
= 2.80 g / 56.11 g/mol
= 0.0499 mol (approximately 0.050 mol)
Since 1 mole of H2SO4 reacts with 2 moles of KOH, the number of moles of H2SO4 required will be double that of KOH:
Number of moles of H2SO4 = 2 * 0.050 mol
= 0.100 mol
Finally, to determine the volume of 0.50 M H2SO4 required, we can use the equation:
Volume (in liters) = number of moles / concentration (in mol/L)
Volume of 0.50 M H2SO4 required = 0.100 mol / 0.50 mol/L
= 0.200 L or 200 mL
Therefore, 200 mL of 0.50 M H2SO4 is required to neutralize completely 2.80 g of KOH.