Hi DrBob, thanks for getting back to me on the earlier related question, but to be honest I am COMPLETELY lost on this! Can you please show me how your worked out the answere using the figures from the question. Thanks.
Question: (a) How much sodium bicarbonate (mass) would have to be used to neutralise 1 L of 18M sulphuric acid
Hi Gille.
I have never answered a question posted by Gille; therefore, you must have posted using another screen name. Thus, I don't know the question you have in mind. Please post the question again and tell me what you don't understand.
Sure, I'd be happy to help you understand how to calculate the amount of sodium bicarbonate needed to neutralize the given amount of sulphuric acid.
To solve this problem, we need to use the balanced chemical equation between sodium bicarbonate (NaHCO3) and sulphuric acid (H2SO4). The balanced equation is:
2 NaHCO3 + H2SO4 -> Na2SO4 + 2 H2O + 2 CO2
Here, we can see that 2 moles of sodium bicarbonate react with 1 mole of sulphuric acid to produce 1 mole of sodium sulfate, 2 moles of water, and 2 moles of carbon dioxide.
Now, let's convert the given concentration of sulphuric acid (18M) to moles per liter. Molarity (M) is defined as moles of solute per liter of solution. So, for 18M sulphuric acid, we have 18 moles of H2SO4 per liter of solution.
To find the amount of sodium bicarbonate needed, we'll use stoichiometry. Stoichiometry allows us to determine the ratio of moles between the reactants and products in a chemical equation.
From the balanced equation, we know that 2 moles of sodium bicarbonate react with 1 mole of sulphuric acid. Therefore, we need half as many moles of sodium bicarbonate as sulphuric acid.
To find the moles of sodium bicarbonate needed, we divide the given moles of sulphuric acid by 2:
moles of NaHCO3 = moles of H2SO4 / 2
Since we're given 1 liter of 18M sulphuric acid, we can calculate the moles of sulphuric acid as follows:
moles of H2SO4 = molarity × volume
moles of H2SO4 = 18 M × 1 L = 18 moles
Now, let's substitute this value into the equation to calculate the moles of sodium bicarbonate needed:
moles of NaHCO3 = 18 moles / 2 = 9 moles
Finally, we convert moles of sodium bicarbonate to mass using its molar mass. The molar mass of sodium bicarbonate is 84.01 g/mol.
mass of NaHCO3 = moles of NaHCO3 × molar mass of NaHCO3
mass of NaHCO3 = 9 moles × 84.01 g/mol
mass of NaHCO3 = 756.09 g
Therefore, to neutralize 1 L of 18M sulphuric acid, we will need approximately 756.09 grams of sodium bicarbonate.
I hope this explanation helps you understand the process of calculating the amount of sodium bicarbonate needed to neutralize sulphuric acid. Let me know if you have any further questions!