Answer:
Option B. decreases by a factor of 100
Explanation:
To obtain the right answer to the question, let us calculate the concentration of the hydronium ion, [H3O+] in both cases.
For pH = 2,
pH = – Log [H3O+]
2 = – Log [H3O+]
–2 = Log [H3O+]
[H3O+] = antilog (–2)
[H3O+] = 1x10^–2M
For pH = 4,
pH = – Log [H3O+]
4 = – Log [H3O+]
–4 = Log [H3O+]
[H3O+] = antilog (–4)
[H3O+] = 1x10^–4M
Now, we shall determine the ratio of [H3O+] in pH 4 to that of pH 2. This is illustrated below:
[H3O+] in pH 4 : [H3O+] in pH 2 => 1x10^–4 / 1x10^–2 = 0.01 = 1x10^–2
From the above, we can thus say that:
[H3O+] in pH 4 = 1/100 [H3O+] in pH 2
Therefore, the [H3O+] decrease by 100
On increasing the pH of the solution from 2 to 4, the Hydronium ion concentration has been decreased by a factor of 100. Thus, option B is correct.
The log of hydronium ion concentration has been calculated as the pH.
pH = -log [Hydronium ion concentration]
Hydronium ion concentration at pH 2:
2 = -log [[tex]\rm H_3O^+[/tex]]
[[tex]\rm H_3O^+[/tex]] = antilog (-2)
[[tex]\rm H_3O^+[/tex]] = 1 [tex]\rm \times\;10^-^2[/tex] M
Hydronium ion concentration at pH 4:
4= -log [[tex]\rm H_3O^+[/tex]]
[[tex]\rm H_3O^+[/tex]] = antilog (-4)
[[tex]\rm H_3O^+[/tex]] = 1 [tex]\rm \times\;10^-^4[/tex] M
The relationship in the hydronium ion at pH 2 and pH 4:
[tex]\rm \dfrac{pH\;2}{pH\;4}\;\times\;\dfrac{10^-^2}{10^-^4}[/tex]
pH 2 = 100 pH 4.
Thus on increasing the pH of the solution from 2 to 4, the Hydronium ion concentration has been decreased by a factor of 100. Thus, option B is correct.
For more information about the hydronium ion concentration, refer to the link:
https://brainly.com/question/12047300
