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Each row of the table below describes an aqueous solution at about [tex]$25^{\circ} C$[/tex]. Complete the table by filling in any missing entries in the second and third columns. Round your entries for [tex]$\left[ H _3 O ^{+}\right]$[/tex] to 2 significant figures and entries for pH to 2 decimal places.

\begin{tabular}{|c|c|c|}
\hline
solution & {[tex]$\left[ H _3 O ^{+}\right]$[/tex]} & pH \\
\hline
A & [tex]$8.3 \times 10^{-10} \, \text{mol/L}$[/tex] & [tex]$\square$[/tex] \\
\hline
B & [tex]$\square \, \text{mol/L}$[/tex] & 7.17 \\
\hline
C & [tex]$8.5 \times 10^{-6} \, \text{mol/L}$[/tex] & [tex]$\square$[/tex] \\
\hline
\end{tabular}

Sagot :

GinnyAnsweridn
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]
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