How much of a 0.10 M H2SO4 solution is required to neutralise 1g of NaOH? Express your answer in cm3.
To determine how much of a 0.10 M H2SO4 solution is required to neutralize 1g of NaOH, we need to use the concept of molarity and the balanced chemical equation between H2SO4 and NaOH.
The balanced equation for the reaction between H2SO4 and NaOH is:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
From the balanced equation, we can see that one mole of H2SO4 reacts with two moles of NaOH. We also know the molarity of the H2SO4 solution, which is 0.10 M.
To find the amount of H2SO4 required, we can use the following equation:
Molarity (M) = Moles (mol) / Volume (L)
Rearranging the equation gives us:
Moles (mol) = Molarity (M) * Volume (L)
Since we want to find the volume in cm3, we need to convert grams to moles using the molar mass of NaOH.
The molar mass of NaOH is:
Na = 22.99 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
So, the molar mass of NaOH is:
22.99 + 16.00 + 1.01 = 40.00 g/mol
Now, we can calculate the moles of NaOH using the equation:
Moles (mol) = Mass (g) / Molar mass (g/mol)
Moles (mol) = 1g / 40.00 g/mol = 0.025 mol
Now that we know the moles of NaOH, we can use the balanced equation to find the moles of H2SO4 required. Since one mole of H2SO4 reacts with two moles of NaOH, we need half the number of moles of H2SO4.
Moles of H2SO4 = 0.025 mol / 2 = 0.0125 mol
Finally, we can calculate the volume of the H2SO4 solution required using the equation:
Volume (L) = Moles (mol) / Molarity (M)
Volume (L) = 0.0125 mol / 0.10 M = 0.125 L
Since we want the volume in cm3, we need to convert from liters to cm3:
1 L = 1000 cm3
Volume (cm3) = 0.125 L * 1000 cm3/L = 125 cm3
Therefore, 125 cm3 of the 0.10 M H2SO4 solution is required to neutralize 1g of NaOH.