Question

calculate the ph from each $h_{3}o^{+}$ concentration.
part 1 of 2
$lefth_{3}o^{+}
ight=6.80\times10^{-4}$ m. be sure your answer has the correct number of significant figures.
$ph = 3.17$
part 2 of 2
$lefth_{3}o^{+}
ight=0.0072$ m. be sure your answer has the correct number of significant figures.
$ph = 2.14$

Explanation:

Part 1 of 2

Step1: Use the pH formula

The formula for pH is \(pH = -\log[H_3O^+]\). Given \([H_3O^+]=6.80\times 10^{-4}\) M.

$$pH=-\log(6.80\times 10^{-4})$$

Using the logarithm property \(\log(ab)=\log a+\log b\), we have \(pH = -(\log(6.80)+\log(10^{-4}))\). Since \(\log(10^{-4})=- 4\) and \(\log(6.80)\approx0.833\), then \(pH=- (0.833 - 4)=3.167\approx3.17\) (three significant figures as \(6.80\) has three significant figures).

Part 2 of 2

Step1: Use the pH formula

Given \([H_3O^+]=0.0072 = 7.2\times 10^{-3}\) M. Using \(pH = -\log[H_3O^+]\), we have \(pH=-\log(7.2\times 10^{-3})\).
Using the logarithm property \(\log(ab)=\log a+\log b\), \(pH=-(\log(7.2)+\log(10^{-3}))\). Since \(\log(10^{-3})=-3\) and \(\log(7.2)\approx0.857\), then \(pH=- (0.857-3)=2.143\approx2.14\) (two significant figures as \(7.2\) has two significant figures).

Answer:

• Part 1 of 2: \(pH = 3.17\)
• Part 2 of 2: \(pH = 2.14\)