Answer :
We start with the definition of pH for an acidic solution. The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration. In mathematical terms:
[tex]$$
\text{pH} = -\log_{10} [H^+]
$$[/tex]
Given the hydrogen ion concentration:
[tex]$$
[H^+] = 2.3 \times 10^{-2} \, \text{mol/L}
$$[/tex]
we substitute this into the pH formula:
[tex]$$
\text{pH} = -\log_{10} (2.3 \times 10^{-2})
$$[/tex]
Recall that the logarithm of a product can be written as the sum of logarithms:
[tex]$$
\log_{10} (2.3 \times 10^{-2}) = \log_{10}(2.3) + \log_{10}(10^{-2})
$$[/tex]
Since:
[tex]$$
\log_{10}(10^{-2}) = -2,
$$[/tex]
the expression becomes:
[tex]$$
\text{pH} = -\left[\log_{10}(2.3) - 2\right] = 2 - \log_{10}(2.3)
$$[/tex]
Evaluating [tex]$\log_{10}(2.3)$[/tex] gives approximately [tex]$0.36$[/tex]. Then,
[tex]$$
\text{pH} \approx 2 - 0.36 = 1.64.
$$[/tex]
Thus, the pH of the solution is approximately [tex]$1.64$[/tex], which corresponds to option D.
[tex]$$
\text{pH} = -\log_{10} [H^+]
$$[/tex]
Given the hydrogen ion concentration:
[tex]$$
[H^+] = 2.3 \times 10^{-2} \, \text{mol/L}
$$[/tex]
we substitute this into the pH formula:
[tex]$$
\text{pH} = -\log_{10} (2.3 \times 10^{-2})
$$[/tex]
Recall that the logarithm of a product can be written as the sum of logarithms:
[tex]$$
\log_{10} (2.3 \times 10^{-2}) = \log_{10}(2.3) + \log_{10}(10^{-2})
$$[/tex]
Since:
[tex]$$
\log_{10}(10^{-2}) = -2,
$$[/tex]
the expression becomes:
[tex]$$
\text{pH} = -\left[\log_{10}(2.3) - 2\right] = 2 - \log_{10}(2.3)
$$[/tex]
Evaluating [tex]$\log_{10}(2.3)$[/tex] gives approximately [tex]$0.36$[/tex]. Then,
[tex]$$
\text{pH} \approx 2 - 0.36 = 1.64.
$$[/tex]
Thus, the pH of the solution is approximately [tex]$1.64$[/tex], which corresponds to option D.
