To complete the table with the given data, we need to use the relationships between the hydronium ion concentration [tex]\(\left[ H_3O^+ \right]\)[/tex] and the pH of the solutions. The relevant formulas we'll apply are:
1. The pH of a solution is given by the formula:
[tex]\[
\text{pH} = -\log_{10} \left[ H_3O^+ \right]
\][/tex]
2. The hydronium ion concentration [tex]\(\left[ H_3O^+ \right]\)[/tex] can be found from the pH using:
[tex]\[
\left[ H_3O^+ \right] = 10^{-\text{pH}}
\][/tex]
Now, let's fill in the missing entries step by step:
### Solution A:
- Given [tex]\(\left[ H_3O^+ \right] = 8.3 \times 10^{-10} \, \text{mol/L}\)[/tex].
- To find the pH for Solution A:
[tex]\[
\text{pH} = -\log_{10} \left( 8.3 \times 10^{-10} \right) \approx 9.08
\][/tex]
So, the completed row for Solution A is:
[tex]\[
\begin{array}{|c|c|c|}
\hline \text{Solution} & \left[ H_3O^+ \right] & \text{pH} \\
\hline A & 8.3 \times 10^{-10} \, \text{mol/L} & 9.08 \\
\hline
\end{array}
\][/tex]
### Solution B:
- Given [tex]\(\text{pH} = 7.17\)[/tex].
- To find [tex]\(\left[ H_3O^+ \right]\)[/tex] for Solution B:
[tex]\[
\left[ H_3O^+ \right] = 10^{-7.17} \approx 6.76 \times 10^{-8} \, \text{mol/L}
\][/tex]
So, the completed row for Solution B is:
[tex]\[
\begin{array}{|c|c|c|}
\hline \text{Solution} & \left[ H_3O^+ \right] & \text{pH} \\
\hline B & 6.76 \times 10^{-8} \, \text{mol/L} & 7.17 \\
\hline
\end{array}
\][/tex]
### Solution C:
- Given [tex]\(\left[ H_3O^+ \right] = 8.5 \times 10^{-6} \, \text{mol/L}\)[/tex].
- To find the pH for Solution C:
[tex]\[
\text{pH} = -\log_{10} \left( 8.5 \times 10^{-6} \right) \approx 5.07
\][/tex]
So, the completed row for Solution C is:
[tex]\[
\begin{array}{|c|c|c|}
\hline \text{Solution} & \left[ H_3O^+ \right] & \text{pH} \\
\hline C & 8.5 \times 10^{-6} \, \text{mol/L} & 5.07 \\
\hline
\end{array}
\][/tex]
Hence, the completed table is:
[tex]\[
\begin{array}{|c|c|c|}
\hline \text{Solution} & \left[ H_3O^+ \right] & \text{pH} \\
\hline A & 8.3 \times 10^{-10} \, \text{mol/L} & 9.08 \\
\hline B & 6.76 \times 10^{-8} \, \text{mol/L} & 7.17 \\
\hline C & 8.5 \times 10^{-6} \, \text{mol/L} & 5.07 \\
\hline
\end{array}
\][/tex]
