Answer :
To determine the pH of a solution with a given hydrogen ion concentration, we use the formula
[tex]$$
\text{pH} = -\log_{10}[H^+]
$$[/tex]
where [tex]$[H^+]$[/tex] is the concentration of hydrogen ions in moles per liter.
Given the concentration
[tex]$$
[H^+] = 2.3 \times 10^{-2} \, \text{mol/L},
$$[/tex]
we substitute into the formula to obtain
[tex]$$
\text{pH} = -\log_{10}(2.3 \times 10^{-2}).
$$[/tex]
Using the properties of logarithms, we break this down as follows:
[tex]$$
\log_{10}(2.3 \times 10^{-2}) = \log_{10}(2.3) + \log_{10}(10^{-2}).
$$[/tex]
It is known that
[tex]$$
\log_{10}(10^{-2}) = -2.
$$[/tex]
Assuming the value for [tex]$\log_{10}(2.3)$[/tex] is approximately [tex]$0.3617$[/tex], we have
[tex]$$
\log_{10}(2.3 \times 10^{-2}) \approx 0.3617 - 2 = -1.6383.
$$[/tex]
Thus, the pH is
[tex]$$
\text{pH} = -(-1.6383) \approx 1.6383.
$$[/tex]
Rounded to two decimal places, the pH is approximately [tex]$1.64$[/tex].
Therefore, the correct answer is:
[tex]$$
\boxed{1.64}
$$[/tex]
[tex]$$
\text{pH} = -\log_{10}[H^+]
$$[/tex]
where [tex]$[H^+]$[/tex] is the concentration of hydrogen ions in moles per liter.
Given the concentration
[tex]$$
[H^+] = 2.3 \times 10^{-2} \, \text{mol/L},
$$[/tex]
we substitute into the formula to obtain
[tex]$$
\text{pH} = -\log_{10}(2.3 \times 10^{-2}).
$$[/tex]
Using the properties of logarithms, we break this down as follows:
[tex]$$
\log_{10}(2.3 \times 10^{-2}) = \log_{10}(2.3) + \log_{10}(10^{-2}).
$$[/tex]
It is known that
[tex]$$
\log_{10}(10^{-2}) = -2.
$$[/tex]
Assuming the value for [tex]$\log_{10}(2.3)$[/tex] is approximately [tex]$0.3617$[/tex], we have
[tex]$$
\log_{10}(2.3 \times 10^{-2}) \approx 0.3617 - 2 = -1.6383.
$$[/tex]
Thus, the pH is
[tex]$$
\text{pH} = -(-1.6383) \approx 1.6383.
$$[/tex]
Rounded to two decimal places, the pH is approximately [tex]$1.64$[/tex].
Therefore, the correct answer is:
[tex]$$
\boxed{1.64}
$$[/tex]
