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What is the pH of a solution containing [tex]$2.3 \times 10^{-2} \, \text{mol/L}$[/tex] of [tex]$H^{+}$[/tex] ions?

A. [tex]\quad-2.3[/tex]
B. [tex]-1.64[/tex]
C. [tex]\quad 2.3[/tex]
D. [tex]\quad 1.64[/tex]

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.

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