Lakes that have been acidified by acid rain can be neutralized by liming, the addition of limestone (CaCO3).
How much limestone in kilograms is required to completely neutralize a 5.8×109 L lake with a pH of 5.6?
thanks
I meant 5.8×10^9 L
To calculate the amount of limestone required to neutralize the lake, we need to determine the initial acidity of the lake based on its pH value, and then calculate the quantity of limestone needed to neutralize that acidity.
The pH scale measures the acidity or alkalinity of a solution, with values below 7 considered acidic. The pH of 5.6 indicates that the lake is slightly acidic.
To neutralize the acid in the lake, we need to calculate the amount of acid present. Acid rain typically contains sulfuric acid (H2SO4), which reacts with calcium carbonate (CaCO3) in limestone.
First, we need to convert the volume of the lake from liters to cubic meters, as the density of limestone is commonly expressed in kilograms per cubic meter.
1 cubic meter (m^3) = 1000 liters (L)
Therefore, the volume of the lake is:
5.8×10^9 L ÷ 1000 = 5.8×10^6 m^3
Next, we need to calculate the acid concentration in the lake using the equation:
[H+] = 10^(-pH)
[H+] represents the concentration of hydrogen ions, which is a measure of the acidity.
[H+] = 10^(-5.6) = 2.51×10^(-6) moles per liter (mol/L)
Now, we can calculate the moles of acid in the lake by multiplying the acid concentration by the lake volume:
Moles of acid = [H+] × Volume = 2.51×10^(-6) mol/L × 5.8×10^9 L = 1.456×10^4 moles
The neutralizing reaction between the acid (H2SO4) and limestone (CaCO3) can be represented as follows:
H2SO4 + CaCO3 → CaSO4 + H2O + CO2
From the balanced equation, we can see that 1 mole of acid reacts with 1 mole of limestone. Therefore, we need the same number of moles of limestone as there are moles of acid in the lake.
The molar mass of limestone (CaCO3) is:
40.08 g/mol (for Ca) + 12.01 g/mol (for C) + 3(16.00 g/mol) (for O) = 100.09 g/mol
To calculate the mass of limestone required, we multiply the moles of acid by the molar mass of limestone:
Mass limestone = Moles of acid × Molar mass = 1.456×10^4 moles × 100.09 g/mol = 1.457×10^6 g
Finally, let's convert the mass of limestone from grams to kilograms:
Mass limestone = 1.457×10^6 g ÷ 1000 = 1.457×10^3 kg
Therefore, approximately 1.457×10^3 kilograms (or 1457 kilograms) of limestone is required to completely neutralize the 5.8×10^9 L lake with a pH of 5.6.
To determine the amount of limestone needed to neutralize a lake, we need to calculate the amount of acid present in the lake and then determine the amount of limestone required to neutralize that acid.
Here's the step-by-step process:
1. Calculate the moles of acid in the lake:
- The pH of the lake is given as 5.6. To find the concentration of hydrogen ions (H+) in moles per liter, we'll use the formula: [H+] = 10^(-pH).
- So, [H+] = 10^(-5.6).
2. Calculate the moles of acid in the entire lake:
- To find the moles of acid in the lake, we multiply the concentration of acid (moles per liter) by the volume of the lake (liters).
- Moles of acid = [H+] * volume of the lake.
3. Convert the moles of acid to the moles of limestone:
- The balanced equation for the reaction between acid and limestone is:
CaCO3 (s) + 2H+ (aq) → Ca2+ (aq) + CO2 (g) + H2O (l).
- From the equation, we can see that 1 mole of limestone reacts with 2 moles of acid.
- Therefore, moles of limestone required = (moles of acid) / 2.
4. Convert the moles of limestone to kilograms:
- The molar mass of CaCO3 (limestone) is 100.09 grams/mole.
- Convert the moles of limestone into grams by multiplying by the molar mass.
- Finally, convert grams to kilograms by dividing by 1000.
Now, let's do the calculations:
Step 1:
[H+] = 10^(-5.6) = 2.51 x 10^(-6) moles/L
Step 2:
Moles of acid = [H+] * Volume of lake
= (2.51 x 10^(-6) moles/L) * (5.8 x 10^9 L)
= 1.4568 x 10^4 moles
Step 3:
Moles of limestone = Moles of acid / 2
= 1.4568 x 10^4 moles / 2
= 7.284 x 10^3 moles
Step 4:
Mass of limestone = Moles of limestone * Molar mass of limestone
= (7.284 x 10^3 moles) * (100.09 g/mole)
= 7.2884256 x 10^5 grams
Finally, to convert grams to kilograms:
Mass of limestone (in kilograms) = 7.2884256 x 10^5 grams / 1000
= 728.84 kilograms
Therefore, approximately 728.84 kilograms of limestone are required to completely neutralize the 5.8×10^9 L lake with a pH of 5.6.
CaCO3 + 2H^+ ==> Ca^+2 + H2O + CO2
What is the H^+?
pH = 5.6; therefore, H^+ = 2.51E-6M
How many mols H^+ do you have? That's M x L = 2.51E-6M x 5.8E9L = 14,569 moles.
How many moles CaCO3 will it take? It will take 1/2 of the H^+; therefore. 14,569/2 = 7,284 moles CaCO3.
How many grams is that?
g = moles x molar mass = ??
Convert that to kg.
Check my numbers.