how many grams of calcium bicarbonate would it take to reach a ph of 7 if i have 10ml 3 molar sulfuric acid?
Ca(HCO3)2 + H2SO4 ==> 2H2O + 2CO2 + CaSO4
I presume your pH 7 part assumes that is the pH at the equivalence point.
mols H2SO4 = M x L = ?
mols Ca(HCO3)2 = mols H2SO4 and you tell that by looking at the coefficients in the balanced equation.
Then grams = mols x molar mass Ca(HCO3)2.
i think you meant carbonate not bicarbonate
To determine the number of grams of calcium bicarbonate needed to reach a pH of 7 with 10 ml of 3 Molar sulfuric acid, we need to understand the chemical reaction between calcium bicarbonate and sulfuric acid.
First, let's look at the balanced chemical equation for the reaction:
Ca(HCO3)2 + H2SO4 → CaSO4 + 2CO2 + 2H2O
From the balanced equation, we can see that 1 mole of calcium bicarbonate (Ca(HCO3)2) reacts with 1 mole of sulfuric acid (H2SO4) to produce 1 mole of calcium sulfate (CaSO4), 2 moles of carbon dioxide (CO2), and 2 moles of water (H2O).
Next, calculate the number of moles of sulfuric acid available:
Moles of sulfuric acid = (volume in liters) × (molarity)
= (10 ml ÷ 1000 ml/L) × 3 M
= 0.03 moles
Since the reaction between calcium bicarbonate and sulfuric acid has a 1:1 mole ratio, the number of moles of calcium bicarbonate required to neutralize the sulfuric acid and reach a pH of 7 is also 0.03 moles.
Now, we need to calculate the molar mass of calcium bicarbonate to convert moles to grams:
Molar mass of calcium bicarbonate (Ca(HCO3)2) = (40.08 g/mol) + (2 × (1.01 g/mol)) + (3 × (16.00 g/mol)) + (2 × (12.01 g/mol))
= 162.10 g/mol
Finally, multiply the number of moles of calcium bicarbonate by its molar mass to obtain the grams required:
Grams of calcium bicarbonate = (moles) × (molar mass)
= 0.03 moles × 162.10 g/mol
= 4.863 grams
Therefore, it would take approximately 4.863 grams of calcium bicarbonate to neutralize the sulfuric acid and reach a pH of 7 when it reacts with 10 ml of 3 Molar sulfuric acid.