Answer :
Let's solve the problem step-by-step.
1. Determining the pH of Beaker \#2:
Given:
- The pH of Beaker \#1 is 2.
Since there is no additional information suggesting a difference in the solutions, we assume that the solution in Beaker \#2 is the same as in Beaker \#1. Therefore, the best prediction for the pH of Beaker \#2 is the same as Beaker \#1:
[tex]\[
\text{pH of Beaker \#2} = 2
\][/tex]
2. Predicting the pH of Beaker \#3:
To predict the pH of Beaker \#3, we need to take into account the dilution. Since 1 mL of the solution from Beaker \#2 is added to 99 mL of water in Beaker \#3, the total volume of Beaker \#3 becomes 100 mL.
- Original solution in Beaker \#2 = 1 mL
- Added water = 99 mL
- Total volume in Beaker \#3 = 1 mL + 99 mL = 100 mL
The concentration of hydrogen ions [tex]\([H^+]\)[/tex] in Beaker \#2 can be found from its pH:
[tex]\[
\text{pH} = -\log_{10}([H^+])
\][/tex]
For Beaker \#2, pH = 2:
[tex]\[
2 = -\log_{10}([H^+])
\][/tex]
Thus:
[tex]\[
[H^+] = 10^{-2} = 0.01 \text{ M}
\][/tex]
When this 1 mL solution with [tex]\([H^+] = 0.01 \text{ M}\)[/tex] is diluted to 100 mL, the concentration of [tex]\([H^+]\)[/tex] in Beaker \#3 is:
[tex]\[
[H^+]_{\text{Beaker \#3}} = \frac{0.01 \, \text{M}}{100} = 0.0001 \, \text{M}
\][/tex]
Now, calculating the pH of Beaker \#3:
[tex]\[
\text{pH of Beaker \#3} = -\log_{10}(0.0001) = -\log_{10}(10^{-4}) = 4.0
\][/tex]
Therefore, the pH of Beaker \#3 is:
[tex]\[
\text{pH of Beaker \#3} = 4.0
\][/tex]
3. Comparing the Concentrations of [tex]$[H^+]$[/tex] in Beaker \#1 and Beaker \#3:
The concentration of [tex]\([H^+]\)[/tex] in Beaker \#1 is:
[tex]\[
[H^+]_{\text{Beaker \#1}} = 10^{-2} = 0.01 \text{ M}
\][/tex]
The concentration of [tex]\([H^+]\)[/tex] in Beaker \#3 is:
[tex]\[
[H^+]_{\text{Beaker \#3}} = 0.0001 \text{ M} = 10^{-4} \text{ M}
\][/tex]
The ratio of concentrations between Beaker \#1 and Beaker \#3 is:
[tex]\[
\frac{[H^+]_{\text{Beaker \#1}}}{[H^+]_{\text{Beaker \#3}}} = \frac{0.01}{0.0001} = 100
\][/tex]
Therefore, the concentration of [tex]\([H^+]\)[/tex] in Beaker \#1 is 100 times greater than in Beaker \#3.
Summary:
- The best prediction for the pH of Beaker \#2 is 2.
- The predicted pH of Beaker \#3 is 4.0.
- The concentration of [tex]\([H^+]\)[/tex] in Beaker \#1 is 100 times greater than in Beaker \#3.
1. Determining the pH of Beaker \#2:
Given:
- The pH of Beaker \#1 is 2.
Since there is no additional information suggesting a difference in the solutions, we assume that the solution in Beaker \#2 is the same as in Beaker \#1. Therefore, the best prediction for the pH of Beaker \#2 is the same as Beaker \#1:
[tex]\[
\text{pH of Beaker \#2} = 2
\][/tex]
2. Predicting the pH of Beaker \#3:
To predict the pH of Beaker \#3, we need to take into account the dilution. Since 1 mL of the solution from Beaker \#2 is added to 99 mL of water in Beaker \#3, the total volume of Beaker \#3 becomes 100 mL.
- Original solution in Beaker \#2 = 1 mL
- Added water = 99 mL
- Total volume in Beaker \#3 = 1 mL + 99 mL = 100 mL
The concentration of hydrogen ions [tex]\([H^+]\)[/tex] in Beaker \#2 can be found from its pH:
[tex]\[
\text{pH} = -\log_{10}([H^+])
\][/tex]
For Beaker \#2, pH = 2:
[tex]\[
2 = -\log_{10}([H^+])
\][/tex]
Thus:
[tex]\[
[H^+] = 10^{-2} = 0.01 \text{ M}
\][/tex]
When this 1 mL solution with [tex]\([H^+] = 0.01 \text{ M}\)[/tex] is diluted to 100 mL, the concentration of [tex]\([H^+]\)[/tex] in Beaker \#3 is:
[tex]\[
[H^+]_{\text{Beaker \#3}} = \frac{0.01 \, \text{M}}{100} = 0.0001 \, \text{M}
\][/tex]
Now, calculating the pH of Beaker \#3:
[tex]\[
\text{pH of Beaker \#3} = -\log_{10}(0.0001) = -\log_{10}(10^{-4}) = 4.0
\][/tex]
Therefore, the pH of Beaker \#3 is:
[tex]\[
\text{pH of Beaker \#3} = 4.0
\][/tex]
3. Comparing the Concentrations of [tex]$[H^+]$[/tex] in Beaker \#1 and Beaker \#3:
The concentration of [tex]\([H^+]\)[/tex] in Beaker \#1 is:
[tex]\[
[H^+]_{\text{Beaker \#1}} = 10^{-2} = 0.01 \text{ M}
\][/tex]
The concentration of [tex]\([H^+]\)[/tex] in Beaker \#3 is:
[tex]\[
[H^+]_{\text{Beaker \#3}} = 0.0001 \text{ M} = 10^{-4} \text{ M}
\][/tex]
The ratio of concentrations between Beaker \#1 and Beaker \#3 is:
[tex]\[
\frac{[H^+]_{\text{Beaker \#1}}}{[H^+]_{\text{Beaker \#3}}} = \frac{0.01}{0.0001} = 100
\][/tex]
Therefore, the concentration of [tex]\([H^+]\)[/tex] in Beaker \#1 is 100 times greater than in Beaker \#3.
Summary:
- The best prediction for the pH of Beaker \#2 is 2.
- The predicted pH of Beaker \#3 is 4.0.
- The concentration of [tex]\([H^+]\)[/tex] in Beaker \#1 is 100 times greater than in Beaker \#3.
