Calculate in cm³ of 0.5mol/dm³ H2SO4 that will neutralize 25cm3 of 0.1mol/dm³ NaOH solution
To calculate the volume of H2SO4 needed to neutralize the NaOH solution, we can use the following balanced chemical equation:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
From the balanced equation, we can see that 2 moles of NaOH react with 1 mole of H2SO4 to produce 2 moles of water.
Given:
Concentration of H2SO4 (C1) = 0.5 mol/dm³
Volume of H2SO4 (V1) = ?? cm³
Concentration of NaOH (C2) = 0.1 mol/dm³
Volume of NaOH (V2) = 25 cm³
First, we need to calculate the number of moles of NaOH in the volume given:
Moles of NaOH (n2) = C2 * V2 = 0.1 mol/dm³ * 25 cm³ = 2.5 mmol
Since 2 moles of NaOH react with 1 mole of H2SO4, we can determine the number of moles of H2SO4 needed:
Moles of H2SO4 (n1) = 1/2 * Moles of NaOH (n2) = 1/2 * 2.5 mmol = 1.25 mmol
Now, we can use the relation between concentration, volume, and moles to find the volume of H2SO4 needed. Rearranging the equation: Concentration (C) = Moles (n) / Volume (V), we have:
V1 = n1 / C1
V1 = 1.25 mmol / 0.5 mol/dm³
V1 = 2.5 cm³
Therefore, 2.5 cm³ of 0.5 mol/dm³ H2SO4 solution is needed to neutralize 25 cm³ of 0.1 mol/dm³ NaOH solution.