How much limestone in kilograms is required to completely neutralize a 3.8 x 109 L lake with a pH of 5.5?
pH 5.5 is what acidity in mol/L?
pH = -log(H^+)
How many mols H^+ in the lake? That's ?mol/L x 3.8E9L = ?mols H^+.
2H^+ + CaCO3 ==> Ca^2+ + CO2 + H2O
? mols H^+ x 1/2 = mols CaCO3.
Convert mols CaCO3 to kg.
To find out the amount of limestone required to neutralize a lake, we need to calculate the amount of acid in the lake and then determine the amount of limestone needed to neutralize that acid.
Step 1: Calculate the amount of acid in the lake
We need to determine the concentration of acid in the lake using the pH value. The pH scale measures the acidity or alkalinity of a solution, with lower values indicating greater acidity.
The formula to calculate the concentration of acid in a solution is:
[H+] = 10^(-pH)
[H+] represents the concentration of hydrogen ions (acid) in moles per liter.
Substituting the given pH value of 5.5 into the formula, we get:
[H+] = 10^(-5.5)
Step 2: Calculate the moles of acid in the lake
To determine the amount of acid in moles, we need to multiply the concentration of acid by the volume of the lake.
Moles of acid = [H+] * Volume of lake
Given the volume of the lake is 3.8 x 10^9 L, the moles of acid can be calculated as:
Moles of acid = (10^(-5.5)) * (3.8 x 10^9)
Step 3: Convert moles of acid to moles of limestone
The reaction between limestone (calcium carbonate, CaCO3) and acid (H+) is as follows:
CaCO3 + 2H+ → Ca2+ + H2O + CO2
From the balanced equation, it can be seen that 1 mole of CaCO3 reacts with 2 moles of H+ (acid). Therefore, we need to convert the moles of acid to moles of limestone using the stoichiometry of the reaction.
Step 4: Calculate the mass of limestone required
The molar mass of limestone (CaCO3) is approximately 100 g/mol. We can use this to convert the moles of limestone to the mass in kilograms.
Mass of limestone (g) = Moles of limestone * Molar mass of limestone
Mass of limestone (kg) = Mass of limestone (g) / 1000
By following these steps, you can calculate the mass of limestone required to neutralize the lake.
To calculate the amount of limestone required to completely neutralize the lake, we need to consider the chemical reaction that occurs when limestone reacts with acid:
CaCO3 + 2H+ -> Ca2+ + H2O + CO2
From the balanced equation, we can see that one mole of CaCO3 reacts with 2 moles of H+, resulting in 1 mole of Ca2+. To find the amount of limestone needed, we'll follow these steps:
Step 1: Calculate the number of moles of H+ in the lake:
Since the lake has a pH of 5.5, we can convert this pH value to the concentration of H+ ions using the equation:
[H+] = 10^(-pH)
[H+] = 10^(-5.5)
[H+] = 3.16 x 10^(-6) mol/L
To convert this to moles, we need to multiply by the volume of the lake:
Number of moles of H+ = [H+] x Volume of the lake
Step 2: Using the stoichiometry of the reaction, determine the number of moles of CaCO3 required:
From the balanced equation, we know that 1 mole of CaCO3 reacts with 2 moles of H+:
Number of moles of CaCO3 = (Number of moles of H+) / 2
Step 3: Convert the moles of CaCO3 to kilograms:
We know that the molar mass of CaCO3 is approximately 100.09 g/mol, which is equivalent to 0.10009 kg/mol.
Mass of CaCO3 = Number of moles of CaCO3 x Molar mass of CaCO3
Let's plug in the given values and calculate the result.
Number of moles of H+ = (3.16 x 10^(-6) mol/L) x (3.8 x 10^9 L) = 1.20 x 10^4 mol
Number of moles of CaCO3 = (1.20 x 10^4 mol) / 2 = 6.00 x 10^3 mol
Mass of CaCO3 = (6.00 x 10^3 mol) x (0.10009 kg/mol) = 600.54 kg
Therefore, approximately 600.54 kilograms of limestone is required to completely neutralize the 3.8 x 10^9 L lake with a pH of 5.5.