How do I calculate how much sodium bicarbonate I need to neutralise 1L of 18M sulphuric acid?
Write the balanced equation.
2NaHCO3 + H2SO4 ==>2H2O + 2CO2 + Na2SO4
Convert 1 L 18M H2SO4 to moles. M x L = moles.
Using the coefficients in the balanced equation, convert moles H2SO4 to moles NaHCO3.
Finally, convert moles NaHCO3 to grams.
grams = moles x molar mass.
Post your work if you get stuck.
Hi Dr Bob
Thanks for getting back but unfortunatley I am completely lost on this! Can you give me the figures please before I go nuts!!
Thanks G
To calculate how much sodium bicarbonate (NaHCO3) you need to neutralize 1L of 18M sulfuric acid (H2SO4), you first need to determine the stoichiometry of the reaction between the two substances. The balanced chemical equation for the reaction is:
2 NaHCO3 + H2SO4 → Na2SO4 + 2 H2O + 2 CO2
From the equation, you can see that for every 2 moles of sodium bicarbonate, you will need 1 mole of sulfuric acid.
Now, let's calculate the number of moles of sulfuric acid in 1L of an 18M solution:
Molarity (M) is defined as moles of solute divided by liters of solution. Rearranging the equation, we have:
Moles of solute = Molarity × Volume
Moles of H2SO4 = 18M × 1L = 18 moles
Since the stoichiometry of the reaction is 2:1, you will need half as many moles of sodium bicarbonate as sulfuric acid. Therefore, you will need:
Moles of NaHCO3 = 18 moles ÷ 2 = 9 moles
To convert moles to grams, you can use the molar mass of sodium bicarbonate, which is:
Na: 22.99 g/mol
H: 1.01 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Molar mass of NaHCO3 = 22.99 + 1.01 + (12.01 + 3 × 16.00) = 84.01 g/mol
Finally, multiply the moles of NaHCO3 by the molar mass to get the mass required:
Mass of NaHCO3 = Moles of NaHCO3 × Molar mass of NaHCO3
= 9 moles × 84.01 g/mol
≈ 756.09 grams
Therefore, you will need approximately 756.09 grams of sodium bicarbonate to neutralize 1L of 18M sulfuric acid.