Neutralization of water containing H2SO4 and HNO3 with CaSO3.
What is the pH in water if the volume is 1000 liter and the concentration of H2SO4 is 0,0013 M and concentration of H2SO4 is 0,0010 M ? And how much of CaCO3 is needed for neutralization of the water?
I believe you have a typo Both are H2SO4. Did you mean one to be HNO3?
Yes one is HNO3
To determine the pH of water after neutralization, we need to consider the reaction between H2SO4 and CaSO3. The balanced chemical equation for this neutralization reaction is as follows:
H2SO4 + CaSO3 -> Ca(HSO4)2
In this reaction, one mole of H2SO4 reacts with one mole of CaSO3 to produce one mole of Ca(HSO4)2.
Given that the volume of water is 1000 liters and the concentration of H2SO4 is 0.0013 M, we need to determine the number of moles of H2SO4 present in the water. This can be calculated using the formula:
moles of H2SO4 = concentration of H2SO4 * volume of water
moles of H2SO4 = 0.0013 M * 1000 L = 1.3 moles
Similarly, the concentration of HNO3 is given as 0.0010 M, and we calculate the number of moles of HNO3:
moles of HNO3 = concentration of HNO3 * volume of water
moles of HNO3 = 0.0010 M * 1000 L = 1.0 moles
Since H2SO4 and HNO3 both react with CaSO3 in a 1:1 ratio, we can assume that the total number of moles of H2SO4 and HNO3 is 1.3 moles + 1.0 moles = 2.3 moles.
Now, to find the pH of the water after neutralization, we need to determine the concentration of hydrogen ions (H+). Since the strong acid H2SO4 and the strong acid HNO3 fully dissociate in water, the concentration of H+ ions is equal to the moles of H2SO4 and HNO3.
Therefore, the concentration of H+ ions = 2.3 moles / 1000 L = 0.0023 M
The negative logarithm of the concentration of H+ ions gives the pH:
pH = -log(H+) = -log(0.0023) ≈ 2.64
So, the pH of the water after neutralization with CaSO3 is approximately 2.64.
Moving on to the second part of your question, to neutralize the water, we need to use CaCO3, not CaSO3. The balanced chemical equation for this reaction is:
CaCO3 + H2SO4 -> CaSO4 + H2O + CO2
From the equation, we can see that one mole of CaCO3 reacts with one mole of H2SO4.
To calculate the amount of CaCO3 needed, we need to know the concentration of H2SO4. You mentioned two concentrations (0.0013 M and 0.0010 M). Assuming you wanted to mention the concentration of one of the acids as HNO3 instead of H2SO4, let's assume the concentration of H2SO4 is 0.0013 M.
Since the volume of water is 1000 liters, the number of moles of H2SO4 would be:
moles of H2SO4 = concentration of H2SO4 * volume of water
moles of H2SO4 = 0.0013 M * 1000 L = 1.3 moles
Since the stoichiometry of the equation shows that one mole of CaCO3 reacts with one mole of H2SO4, we need 1.3 moles of CaCO3 to neutralize the 1.3 moles of H2SO4 present in the water.
Therefore, you would need 1.3 moles of CaCO3 to neutralize the water. To convert this into grams, you need to know the molar mass of CaCO3, which is approximately 100.09 g/mol.
Mass of CaCO3 = moles of CaCO3 * molar mass of CaCO3
Mass of CaCO3 = 1.3 moles * 100.09 g/mol ≈ 130.12 grams
So, you would need approximately 130.12 grams of CaCO3 to neutralize the water.