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Chapter 18: Problem 76

The CDC (Centers for Disease Control and Prevention) published a reference blood lead level (BLL), which is based on the BLL distribution among children. It is currently \(5 \mu \mathrm{p} / \mathrm{dL} .\) (a) What is the molarity of an aqueous solution with this concentration? (b) Express this concentration in ppb.

Short Answer

Expert verified
(a) The molarity of an aqueous solution with a BLL of 5 μg/dL is \(2.42 \times 10^{-7}\) M. (b) The concentration expressed in ppb is 50 ppb.

Step by step solution

01

Convert BLL to grams per liter (g/L)

BLL is given as 5 μg/dL. We will convert this to g/L: BLL = 5 μg/dL = 5 * \(1 \times 10^{-6}\) g/dL = \(5 \times 10^{-6}\) g/dL Now, convert deciliter (dL) to liter (L): BLL = \(5 \times 10^{-6}\) g/dL * \(\dfrac{1}{0.1}\) L/dL = \(5 \times 10^{-5}\) g/L
02

Calculate moles of Pb

Use the molar mass of lead (Pb) to calculate the moles of Pb: Moles of Pb = \(\dfrac{5 \times 10^{-5}\text{g}}{207.2 \text{g/mol}}\)
03

Calculate the molarity

Now, divide the moles of Pb by the volume (in liters) to get the molarity: Molarity = \(\dfrac{5 \times 10^{-5}\text{g}}{207.2 \text{g/mol} \times 1\text{L}}\) Molarity = \(2.42 \times 10^{-7}\) M #a. Converting the concentration to ppb#
04

Determine the molar ratio of Pb to the solution

We have already calculated the molarity of Pb in the solution, which is the ratio of moles of Pb to liters of the solution. Now, we need to find the concentration ratio: Concentration ratio of Pb = \(\dfrac{2.42 \times 10^{-7}\text{mol}}{1\text{L}}\)
05

Calculate the mass concentration of Pb in the solution (g/L)

To obtain the mass concentration in g/L, we multiply the concentration ratio by the molar mass of Pb: Mass concentration of Pb = \(2.42 \times 10^{-7}\text{mol} \times 207.2\text{g/mol}\) / 1 L = \(5 \times 10^{-5}\) g/L
06

Convert the mass concentration to ppb

Now, we will convert the mass concentration from g/L to ppb (parts per billion): 1 g/L = \(1 \times 10^{9}\) ppb So, BLL in ppb = \(5 \times 10^{-5}\) g/L * \(1 \times 10^{9}\) ppb/g = \(50\) ppb Therefore, (a) the molarity of an aqueous solution with a BLL of 5 μg/dL is \(2.42 \times 10^{-7}\) M, and (b) the concentration expressed in ppb is 50 ppb.

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