McGraw Hill Glencoe Algebra 2, 2012
MH
McGraw Hill Glencoe Algebra 2, 2012 View details
6. Common Logarithms
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Exercise 66 Page 497

Practice makes perfect
a

Let's analyze the given formula.

pH=-log H The variable H is the hydrogen ion concentration. In our case it is equal to 1.25* 10^(- 11). We know that the pH of the arsenic level should be less than 9.5. Let's check it out by substituting 1.25* 10^(- 11) for H and solve it for pH.

pH=-log H
pH=-log( 1.25* 10^(- 11))
pH≈ 10.9

Since 10.9 is greater than 9.5, the environmental engineer should be worried about too high an arsenic content.

b

The safe standard for arsenic is 0.025 parts per million (ppm). The environmental engineer finds 1 milligram of arsenic in a 3 liter sample. Since 1 liter of water is around 1 kilogram, let's find out the ratio of arsenic in the tested water.

1 mg/3 kg=1/3 mg/kg≈ 0.33 mg/kg Since 1 ppm is 1 mg/kg, we get that in our case the level of arsenic is about 0.33 ppm. This value is greater than the safe standard for arsenic. Therefore, the tested well is not safe.
c

Let's analyze the given formula.

pH=-log H H is the hydrogen ion concentration. We are asked to find the hydrogen ion concentration that meets the troublesome pH level of 9.5. Let's substitute 9.5 for pH into the given formula and solve it for H.

pH=-log H
9.5=-log H
- 9.5=log H

To solve the equation for H, we will use the definition of a logarithm. Notice also that log x is equal to log_(10)x for all x. y=log_b x ⇔ x= b^y This definition tells us how to rewrite the logarithm equivalent of y as an exponential equation. The argument x is equal to b raised to the power of y. Let's do it! - 9.5=log_(10) H ⇔ H= 10^(- 9.5) Finally, using a calculator, we get that H=10^(- 9.5)≈ 3.16* 10^(- 10) ppm. Therefore, the hydrogen ion concentration around 3.16* 10^(- 10) ppm meets the troublesome pH level of 9.5.