To solve this problem, let's first understand the relationship between the amount of hydrogen (in moles per liter, denoted as [tex]\(x\)[/tex]) and the pH of the solution.
From the information provided in the table, we have the following pairs of values for the amount of hydrogen ([tex]\(x\)[/tex]) and the pH ([tex]\(f(x)\)[/tex]):
[tex]\[
\begin{array}{|c|c|}
\hline
\text{Amount of Hydrogen, } x \text{ (in moles per liter)} & \text{pH, } f(x) \\
\hline
\frac{1}{10} & 1 \\
\frac{1}{100} & 2 \\
\frac{1}{1000} & 3 \\
\frac{1}{10000} & 4 \\
\frac{1}{100000} & 5 \\
\hline
\end{array}
\][/tex]
To identify the relationship between [tex]\(x\)[/tex] and [tex]\(f(x)\)[/tex], we notice a pattern. The pH value is derived from the amount of hydrogen [tex]\(x\)[/tex] by using the following formula:
[tex]\[
\text{pH} = -\log_{10}(\text{amount of hydrogen in moles per liter})
\][/tex]
Let's apply this formula to each entry in the table:
1. For [tex]\(x = \frac{1}{10} = 10^{-1}\)[/tex]:
[tex]\[
\text{pH} = -\log_{10}\left(10^{-1}\right) = -(-1) = 1
\][/tex]
2. For [tex]\(x = \frac{1}{100} = 10^{-2}\)[/tex]:
[tex]\[
\text{pH} = -\log_{10}\left(10^{-2}\right) = -(-2) = 2
\][/tex]
3. For [tex]\(x = \frac{1}{1000} = 10^{-3}\)[/tex]:
[tex]\[
\text{pH} = -\log_{10}\left(10^{-3}\right) = -(-3) = 3
\][/tex]
4. For [tex]\(x = \frac{1}{10000} = 10^{-4}\)[/tex]:
[tex]\[
\text{pH} = -\log_{10}\left(10^{-4}\right) = -(-4) = 4
\][/tex]
5. For [tex]\(x = \frac{1}{100000} = 10^{-5}\)[/tex]:
[tex]\[
\text{pH} = -\log_{10}\left(10^{-5}\right) = -(-5) = 5
\][/tex]
From these calculations, we have verified that for each given amount of hydrogen [tex]\(x\)[/tex], the corresponding pH is indeed [tex]\(-\log_{10}(x)\)[/tex]. Therefore, the general relationship between the pH and the amount of hydrogen (in moles per liter) can be written as:
[tex]\[
\text{pH} = -\log_{10}(\text{amount of hydrogen in moles per liter})
\][/tex]
This relationship is consistent with the values provided in the table.
