Question

The concentrations of arsenic (As) and selenium (Se) in a drinking water well were 2.0 and 3.8 ppb. (a) Convert arsenic concentration into ppm and mg/L, (b) Convert selenium concentration into molarity (M) and micro-molarity (mM), (c) Have the concentrations exceeded the maximum contaminant level (MCL) of 50 mg/L for both elements? The atomic weight of Se is 79.

Answer 1

Answer on Question #58199 – Chemistry – Other

Task:

The concentrations of arsenic (As) and selenium (Se) in a drinking water well were 2.0 and 3.8 ppb. (a) Convert arsenic concentration into ppm and mg/L, (b) Convert selenium concentration into molarity (M) and micro-molarity (mM), (c) Have the concentrations exceeded the maximum contaminant level (MCL) of 50 mg/L for both elements? The atomic weight of Se is 79.

Solution:

$$C(As, ppb) = 2.0\,ppb;$$ $$C(Se, ppb) = 3.8\,ppb.$$

(a) Convert arsenic concentration into ppm and mg/L:

ppm and ppb are defined as:

$$1\,ppm = 1/10^6 = 10^{-6};$$ $$1\,ppb = 1/10^9 = 10^{-9}.$$

So

$$1\,ppm = 1000\,ppb$$ $$x_{ppm} = x_{ppb} / 1000.$$

Then, the concentrations of arsenic (As) and selenium (Se) into ppm:

$$C(As, ppm) = C(As, ppb) / 1000 = 2.0\,ppb / 1000 = 0.0020\,ppm;$$ $$C(Se, ppm) = C(Se, ppb) / 1000 = 3.8\,ppb / 1000 = 0.0038\,ppm.$$

ppb and mg/l are defined as:

$$1000\,g(H_2O) = 1000\,ml(H_2O) = 1\,l(H_2O), \text{ because the } \rho(H_2O) = 1\,g/ml;$$ $$1\,g = 10^{-3}\,mg;$$ $$1\,ppb = 1/10^9 = 10^{-9} = \frac{10^{-6}\,g}{1000\,g} = \frac{10^{-6}\,g}{1000\,ml} = \frac{10^{-3}\,mg}{1\,l} = 1 \times 10^{-3}\,mg/l.$$

So

$$1\,mg/l = 1000 \times 1\,ppb.$$ $$x_{mg/l} = x_{ppb} \times 10^{-3}.$$

Then, the concentrations of arsenic (As) and selenium (Se) into mg/l:

$$C(As, mg/l) = C(As, ppb) \times 10^{-3} = 2.0\,ppb \times 10^{-3} = 0.0020\,mg/l;$$ $$C(Se, mg/l) = C(Se, ppb) \times 10^{-3} = 3.8\,ppb \times 10^{-3} = 0.0038\,mg/l.$$

**Answer:**

$$C(As) = 2.0\,ppb = 0.0020\,ppm = 0.0020\,mg/l;$$

b) Convert selenium concentration into molarity (M) and micro-molarity (mM):

The atomic weight of Se is 79.

$$M(Se) = 79 \frac{g}{mol} = 79 \times 10^{3} \frac{mg}{l}.$$

The molarity of a solution is calculated by taking the moles of solute and dividing by the liters of solution:

$$Molarity = \frac{moles\ of\ solute}{liters\ of\ solution}.$$

Then,

$$Molarity(Se) = \frac{n(Se)}{V} = \frac{m(Se)}{M(Se) \times V} = \frac{C(Se, mg/l)}{M(Se)}.$$ $$Molarity(Se) = \frac{0.0038\,mg/l}{79 \times 10^{3} \,mg/mol} = 4.8 \times 10^{-8} \,mol/l.$$ $$Micro-molarity(Se) = Molarity(Se) \times 10^{6} = 4.8 \times 10^{-8} \,M \times 10^{6} = 0.048 \,mM.$$

**Answer:**

$$Molarity(Se) = 4.8 \times 10^{-8} \,mol/l = 4.8 \times 10^{-8} \,M;$$ $$Micro-molarity(Se) = 0.048 \,mM.$$

(c) Have the concentrations exceeded the maximum contaminant level (MCL) of $50\,\mathrm{mg/L}$ for both elements?

$$C(Se, mg/l) = 0.0038\,mg/l < \text{the maximum contaminant level (MCL)} = 50\,mg/l.$$ $$C(As, mg/l) = 0.0020\,mg/l < \text{the maximum contaminant level (MCL)} = 50\,mg/l.$$ $$\sum C(As) + C(Se) = 0.0020\,mg/l + 0.0038\,mg/l = 0.0058\,mg/l.$$ $$\sum C(As) + C(Se) = 0.0058\,mg/l < MCL = 50\,mg/l.$$

Since the amount concentrations of selenium and arsenic do not exceed the maximum permissible concentration of arsenic (MCL) of $50\,\mathrm{mg/l}$, it can be said that the level content of both the metal in the water corresponds to the norm.

**Answer:**

The content of selenium and arsenic do not exceed the maximum permissible concentration of arsenic (MCL) of $50\,\mathrm{mg/l}$.

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