Question

Grade: DETERMINATION OF PHOSPHATE IN WATER 1. Option What is the concentration of phosphate in the original lake water sample if the unknown lake water sample in the option had been diluted by taking 25.0 mL of lake water and diluting it to the indicated final volume? 21.comgy 2. Option Water treatment plants must reduce phosphate levels to 3.2 micromoles PO 3-/L before the water is considered drinkable. If water from Lake Pontchartrain is determined to have a phosphate concentration of Zmg PO43-/L, does it meet this allowable standard for drinking water? Show your work! 3. Option Suppose that absorbances of several solutions were measured at nm. What is the frequency of this wavelength? What is the energy of this wavelength? (If you have forgotter relationship between wavelength, frequency, and energy, refer to module 7A in your lecture handbook. h = 6.626 x 10-34 J.)

Answer 1

1. Concentration of phosphate in the diluted lake water sample = 21.6 mg/L.

Volume of the diluted sample = 600.0 mL.

Volume of lake water sample taken = 25.0 mL.

Dilution factor (DF) = (600.0 mL)/(25.0 mL)

= 24

Concentration of phosphate in the original lake water = (21.6 mg/L)*(DF)

= (21.6 mg/L)*(24)

= 518.4 mg/L (ans).

2. Determined phosphate concentration in Lake Pontchartrain = 0.372 mg PO₄^3-/L.

The atomic masses are

P: 31 u

O: 16 u

Gram molar mass of PO₄^3- = (1*31 + 4*16) g/mol

= 95 g/mol

Concentration of phosphate = 0.372 mg/L

= (0.372 mg/L)/(95 g/mol)

= [(0.372 mg/L)*(1 g/1000 mg)]/(95 g/mol)

= (0.000372 g/L)/(95 g/mol)

= 3.916*10^-6 mol/L

= (3.916*10^-6 mol/L)*(1.0*10⁶ micromoles)/(1 mole)

= 3.916 micromoles/L

≈ 3.9 micromoles/L

The concentration of phosphate is higher than the desired phosphate concentration of 3.2 micromoles PO₄^3-/L (ans).

3. Wavelength of the radiation = 620 nm

The frequency and wavelength of the radiation is related by the relation

ν*λ = c

where c = 2.998*10⁸ m.s^-1.

Plug in values and obtain

ν*(620 nm) = 2.998*10⁸ m.s^-1

======> ν*(620 nm)*(1 m)/(1.0*10⁹ nm) = 2.998*10⁸ m.s^-1

======> ν*(6.2*10^-7 m) = 2.998*10⁸ m.s^-1

======> ν = (2.998*10⁸ m.s^-1)/(6.2*10^-7 m)

======> ν = 4.835*10¹⁴ s^-1 ≈ 4.83*10¹⁴ s^-1

The frequency of the radiation = 4.83*10¹⁴ s^-1 = 4.83*10¹⁴ Hz (ans).

The energy of the radiation is given by the relation

E = hν

= (6.626*10^-34 J.s)*(4.83*10¹⁴ s^-1)

= 3.200358*10^-19 J

≈ 3.20*10^-19 J (ans).