Sure! Let's solve the problem step-by-step.
The pH of a solution is a measure of its hydrogen ion concentration [tex]\([H^+]\)[/tex], defined by the relationship:
[tex]\[ \text{pH} = -\log_{10}([H^+]) \][/tex]
When the pH of the solution decreases, it means that the hydrogen ion concentration [tex]\([H^+]\)[/tex] increases.
Given the problem:
- The pH decreases by 2.0
We need to determine by what factor the hydronium ion concentration changes when the pH decreases by 2.0.
1. Understanding the relationship between pH and hydronium ion concentration:
The change in [tex]\( [H^+] \)[/tex] due to a change in pH can be determined by:
[tex]\[ \Delta \text{pH} = -\log_{10}([H^+]_{\text{new}}/[H^+]_{\text{initial}}) \][/tex]
2. Find the factor by which [tex]\( [H^+] \)[/tex] changes:
Given that the decrease in pH is [tex]\( 2.0 \)[/tex]:
[tex]\[
\Delta \text{pH} = 2.0 = -\log_{10}\left(\frac{[H^+]_{\text{new}}}{[H^+]_{\text{initial}}}\right)
\][/tex]
Therefore,
[tex]\[
\log_{10}\left(\frac{[H^+]_{\text{new}}}{[H^+]_{\text{initial}}}\right) = -2.0
\][/tex]
3. Exponentiating to solve for [tex]\([H^+]_{\text{new}}/[H^+]_{\text{initial}}\)[/tex]:
[tex]\[
\frac{[H^+]_{\text{new}}}{[H^+]_{\text{initial}}} = 10^{-(-2.0)} = 10^{2.0} = 100
\][/tex]
Therefore, the hydronium ion concentration increases by a factor of 100.
So the correct answer is:
- increases to 100 times the original concentration
This means that the hydronium ion concentration increases to 100 times its initial value when the pH of the solution decreases by 2.
