Question

calculate the ph from each $h_{3}o^{+}$ concentration.
your answer
part 1 of 2
$h_{3}o^{+}=6.80×10^{-4}m$. be sure your answer has the correct number of significant figures.
$ph=3.17$
part 2 of 2
$h_{3}o^{+}=0.0072m$. be sure your answer has the correct number of significant figures.
$ph=2.14$

Explanation:

Step1: Recall the pH formula

The formula for pH is \(pH = -\log_{10}[H_3O^+]\).

Step2: Calculate pH for \([H_3O^+]=6.80\times 10^{-4}\ M\)

Substitute \([H_3O^+]=6.80\times 10^{-4}\) into the formula:
\(pH = -\log(6.80\times 10^{-4})\)
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 \(\log(6.80)\approx0.833\)
\(pH=- (0.833 - 4)=3.167\approx3.17\)

Step3: Calculate pH for \([H_3O^+]=0.0072\ M = 7.2\times 10^{-3}\ M\)

Substitute \([H_3O^+]=7.2\times 10^{-3}\) into the formula:
\(pH = -\log(7.2\times 10^{-3})\)
Using the logarithm property \(\log(ab)=\log a+\log b\), we have \(pH=-(\log(7.2)+\log(10^{-3}))\)
Since \(\log(10^{-3})=-3\) and \(\log(7.2)\approx0.857\)
\(pH=- (0.857-3)=2.143\approx2.14\)

Answer:

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