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Chapter 4: Problem 114

The U.S. standard for arsenate in drinking water requires that public water supplies must contain no greater than 10 parts per billion \((\mathrm{ppb})\) arsenic. If this arsenic is present as arsenate, AsO \(_{4}^{3-},\) what mass of sodium arsenate would be present in a 1.00 -L sample of drinking water that just meets the standard? Parts per billion is defined on a mass basis as $$\mathrm{ppb}=\frac{\text { g solute }}{\mathrm{g} \text { solution }} \times 10^{9}$$

Short Answer

Expert verified
The mass of sodium arsenate present in a 1.00-L sample of drinking water that just meets the standard for arsenate concentration (10 ppb) is \(10^{-5}\;\text{g}\).

Step by step solution

01

Calculate the mass of the solution

First, we need to find the mass of the 1.00-L water sample. The density of water is approximately 1 g/mL or 1 g/cm³. Since 1 L = 1000 mL = 1000 cm³ , the mass of the water sample is: Mass of water = Volume x Density Mass of water = 1.00 L x 1 g/mL x 1000 mL/L Mass of water = 1000 g
02

Set up the equation for ppb

Now, we will use the equation for ppb: ppb = \(\frac{\text{g solute}}{\text{g solution}} \times 10^9\) We know the allowed ppb for arsenate is 10, the mass of the solute (sodium arsenate) is unknown, and the mass of the solution is 1000g. Substitute the known values into the equation: 10 = \(\frac{\text{g solute}}{1000 \text{ g}} \times 10^9\)
03

Solve for mass of sodium arsenate

Now, let's solve for the mass of sodium arsenate (g solute): g solute = \(\frac{10 \times 1000 \text{ g}}{10^9}\) g solute = \(\frac{10^4 \text{ g}}{10^9}\) g solute = \(10^{-5} \text{ g}\)
04

Calculate mass of sodium arsenate (AsO\(_{4}^{3-}\)) in the 1.00-L sample

Since we have the mass of sodium arsenate (AsO\(_{4}^{3-}\)) in grams, we can now calculate the mass of sodium arsenate present in the 1.00-L sample: Mass of sodium arsenate = \(10^{-5}\;\text{g}\) Therefore, the mass of sodium arsenate present in a 1.00-L sample of drinking water that just meets the standard is \(10^{-5}\;\text{g}\).

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Most popular questions from this chapter

Indicate the concentration of each ion present in the solution formed by mixing (a) 42.0 \(\mathrm{mL}\) of 0.170 \(\mathrm{M} \mathrm{NaOH}\) with 37.6 \(\mathrm{mL}\) of \(0.400 \mathrm{M} \mathrm{NaOH},(\mathbf{b}) 44.0 \mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) with 25.0 \(\mathrm{mL}\) of \(0.150 \mathrm{M} \mathrm{KCl},\) (c) 3.60 \(\mathrm{g} \mathrm{KCl}\) in 75.0 \(\mathrm{mL}\) of 0.250\(M \mathrm{CaCl}_{2}\) solution. Assume that the volumes are additive.

Suppose you have a solution that might contain any or all of the following cations: \(\mathrm{Ni}^{2+}, \mathrm{Ag}^{+}, \mathrm{Sr}^{2+},\) and \(\mathrm{Mn}^{2+} .\) Addition of HCl solution causes a precipitate to form. After filtering off the precipitate, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is added to the resulting solution and another precipitate forms. This is filtered off, and a solution of \(\mathrm{NaOH}\) is added to the resulting solution. No precipitate is observed. Which ions are present in each of the precipitates? Which of the four ions listed above must be absent from the original solution?

The concept of chemical equilibrium is very important. Which one of the following statements is the most correct way to think about equilibrium? (a) If a system is at equilibrium, nothing is happening. (b) If a system is at equilibrium, the rate of the forward reaction is equal to the rate of the back reaction. (c) If a system is at equilibrium, the product concentration is changing over time. Section 4.1\(]\)

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