Answer :
To determine the pH, we use the formula
[tex]$$
\text{pH} = - \log_{10} \left([H^+]\right).
$$[/tex]
Given that the concentration of hydrogen ions is
[tex]$$
[H^+] = 2.3 \times 10^{-2} \text{ mol/L},
$$[/tex]
we substitute this value into the formula:
[tex]$$
\text{pH} = - \log_{10} \left(2.3 \times 10^{-2}\right).
$$[/tex]
First, recognize that
[tex]$$
\log_{10} \left(2.3 \times 10^{-2}\right) = \log_{10}(2.3) + \log_{10}(10^{-2}).
$$[/tex]
Since
[tex]$$
\log_{10}(10^{-2}) = -2,
$$[/tex]
we have
[tex]$$
\log_{10} \left(2.3 \times 10^{-2}\right) \approx \log_{10}(2.3) - 2.
$$[/tex]
Calculating [tex]$\log_{10}(2.3)$[/tex], we obtain approximately [tex]$0.36$[/tex]. Hence,
[tex]$$
\log_{10} \left(2.3 \times 10^{-2}\right) \approx 0.36 - 2 = -1.64.
$$[/tex]
Finally, taking the negative of this value gives
[tex]$$
\text{pH} = -(-1.64) = 1.64.
$$[/tex]
Thus, the pH of the solution is approximately [tex]$1.64$[/tex], which corresponds to option D.
[tex]$$
\text{pH} = - \log_{10} \left([H^+]\right).
$$[/tex]
Given that the concentration of hydrogen ions is
[tex]$$
[H^+] = 2.3 \times 10^{-2} \text{ mol/L},
$$[/tex]
we substitute this value into the formula:
[tex]$$
\text{pH} = - \log_{10} \left(2.3 \times 10^{-2}\right).
$$[/tex]
First, recognize that
[tex]$$
\log_{10} \left(2.3 \times 10^{-2}\right) = \log_{10}(2.3) + \log_{10}(10^{-2}).
$$[/tex]
Since
[tex]$$
\log_{10}(10^{-2}) = -2,
$$[/tex]
we have
[tex]$$
\log_{10} \left(2.3 \times 10^{-2}\right) \approx \log_{10}(2.3) - 2.
$$[/tex]
Calculating [tex]$\log_{10}(2.3)$[/tex], we obtain approximately [tex]$0.36$[/tex]. Hence,
[tex]$$
\log_{10} \left(2.3 \times 10^{-2}\right) \approx 0.36 - 2 = -1.64.
$$[/tex]
Finally, taking the negative of this value gives
[tex]$$
\text{pH} = -(-1.64) = 1.64.
$$[/tex]
Thus, the pH of the solution is approximately [tex]$1.64$[/tex], which corresponds to option D.
