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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. -2.3
B. -1.64
C. 2.3
D. 1.64

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]

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