The maximum allowable concentration of Pb2+ ions in drinking water is 0.05 ppm (i.e., 0.05 g of Pb2+ in 1 million grams of water
1 answer:
Answer:
a. 0,049g of Pb²⁺ in 1 million grams of water.
b. No
Explanation:
The equilibrium of PbCO₃ in water is:
PbCO₃ ⇄ Pb²⁺ + CO₃²⁻
Where ksp is defined as:
ksp = [Pb²⁺] [CO₃²⁻] <em>(1)</em>
The initial concentration of PbCO₃ in molarity is:
×
= 0,0374M
a. The concentrations in equilibrium are:
[PbCO₃} = 0,0374M-x
[Pb²⁺] = x
[CO₃²⁻] = x
Replacing in (1)
3,3x10⁻¹⁴ = x²
x = 1,82x10⁻⁷M
Thus, [Pb²⁺] = 1,82x10⁻⁷M, this value in g of Pb²⁺ per million grams of water is:
1,82x10⁻⁷M××
= <em>0,049 g of Pb²⁺ in 1 million grams of water</em>
As this value is lower than the maximum allowable concentration of Pb²⁺, <em>the concentration doesn't exceed this maximum value</em>
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Answer:
0.03g/mL
Explanation:
Given parameters include:
Five μL of a 10-to-1 dilution of a sample; This implies the Volume of dilute sample is given as 5 μL
Dilution factor = 10-to-1
The absorbance at 595 nm was 0.78
Mass of the diluted sample = 0.015 mg
We need to first determine the concentration of the diluted sample which is required in calculating the protein concentration of the original solution.
So, to determine the concentration of the diluted sample, we have:
concentration of diluted sample =
= (where ∝ was use in place of μ in the expressed fraction)
= 0.003 mg/μL
The dilution of the sample is from 10-to-1 indicating that the original concentration is ten times higher; as such the protein concentration of the original solution can be calculated as:
protein concentration of the original solution = 10 × concentration of the diluted sample.
= 10 × 0.003 mg/μL
= 0.03 mg/μL
= 0.03g/mL
Hence, the protein concentration of the original solution is known to be 0.03g/mL