Answer :
To complete the chart, we'll need to determine the missing values for each row. We can utilize the relationships between pH, pOH, [tex]\([\text{H}^+]\)[/tex], and [tex]\([\text{OH}^-]\)[/tex] to find all the necessary values.
1. Row 1: Given pH = 3.5
[tex]\[ \text{pH} + \text{pOH} = 14 \Rightarrow \text{pOH} = 14 - \text{pH} = 14 - 3.5 = 10.5 \][/tex]
Next, we'll use [tex]\([\text{H}^+] = 10^{-\text{pH}}\)[/tex].
[tex]\[ [\text{H}^+] = 10^{-3.5} \][/tex]
This can be approximated but is commonly used:
[tex]\[ [\text{H}^+] \approx 3.16 \times 10^{-4} \][/tex]
Lastly, using the relationship [tex]\([\text{H}^+][\text{OH}^-] = 1 \times 10^{-14}\)[/tex]:
[tex]\[ [\text{OH}^-] = \frac{1 \times 10^{-14}}{[\text{H}^+]} = \frac{1 \times 10^{-14}}{3.16 \times 10^{-4}} \approx 3.16 \times 10^{-11} \][/tex]
- pH = 3.5, pOH = 10.5, [\text{H}^+] = 3.16 \times 10^{-4}, [\text{OH}^-] \approx 3.16 \times 10^{-11}
2. Row 2: Given pOH = 2.9
[tex]\[ \text{pH} + \text{pOH} = 14 \Rightarrow \text{pH} = 14 - \text{pOH} = 14 - 2.9 = 11.1 \][/tex]
To find [tex]\([\text{H}^+]\)[/tex]:
[tex]\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-11.1} \][/tex]
This can be approximated as:
[tex]\[ [\text{H}^+] \approx 7.94 \times 10^{-12} \][/tex]
Lastly, using [tex]\([\text{H}^+][\text{OH}^-] = 1 \times 10^{-14}\)[/tex] again:
[tex]\[ [\text{OH}^-] = 10^{-2.9} \][/tex]
This value can be calculated easily as:
[tex]\[ [\text{OH}^-] \approx 1.26 \times 10^{-3} \][/tex]
- pH = 11.1, pOH = 2.9, [\text{H}^+] \approx 7.94 \times 10^{-12}, [\text{OH}^-] \approx 1.26 \times 10^{-3}
3. Row 3: Given [tex]\([\text{H}^+] = 4.3 \times 10^{-4}\)[/tex]
Using [tex]\(\text{pH} = -\log [\text{H}^+]\)[/tex]:
[tex]\[ \text{pH} = -\log (4.3 \times 10^{-4}) \approx 3.37 \][/tex]
Then, [tex]\(\text{pH} + \text{pOH} = 14\)[/tex]:
[tex]\[ \text{pOH} = 14 - 3.37 = 10.63 \][/tex]
Subsequently, [tex]\([\text{OH}^-]\)[/tex] can be calculated:
[tex]\[ [\text{OH}^-] = \frac{1 \times 10^{-14}}{4.3 \times 10^{-4}} \approx 2.33 \times 10^{-11} \][/tex]
- pH \approx 3.37, pOH \approx 10.63, [\text{H}^+] = 4.3 \times 10^{-4}, [\text{OH}^-] \approx 2.33 \times 10^{-11}
4. Row 4: Given [tex]\([\text{OH}^-] = 5.7 \times 10^{-8}\)[/tex]
Using [tex]\(\text{pOH} = -\log [\text{OH}^-]\)[/tex]:
[tex]\[ \text{pOH} = -\log (5.7 \times 10^{-8}) \approx 7.24 \][/tex]
Then, [tex]\(\text{pOH} + \text{pH} = 14\)[/tex]:
[tex]\[ \text{pH} = 14 - 7.24 = 6.76 \][/tex]
Subsequently, [tex]\([\text{H}^+]\)[/tex] can be calculated:
[tex]\[ [\text{H}^+] = \frac{1 \times 10^{-14}}{5.7 \times 10^{-8}} \approx 1.75 \times 10^{-7} \][/tex]
- pH \approx 6.76, pOH \approx 7.24, [\text{H}^+] \approx 1.75 \times 10^{-7}, [\text{OH}^-] = 5.7 \times 10^{-8}
Completed table:
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
pH & pOH & {[tex]$\left[ \text{H} ^{+}\right]$[/tex]} & {[tex]$\left[ \text{OH} ^{-1}\right.$[/tex]} \\
\hline
3.5 & 10.5 & [tex]$3.16 \times 10^{-4}$[/tex] & [tex]$3.16 \times 10^{-11}$[/tex] \\
\hline
11.1 & 2.9 & [tex]$7.94 \times 10^{-12}$[/tex] & [tex]$1.26 \times 10^{-3}$[/tex] \\
\hline
3.37 & 10.63 & [tex]$4.3 \times 10^{-4}$[/tex] & [tex]$2.33 \times 10^{-11}$[/tex] \\
\hline
6.76 & 7.24 & [tex]$1.75 \times 10^{-7}$[/tex] & [tex]$5.7 \times 10^{-8}$[/tex] \\
\hline
\end{tabular}
\end{center}
1. Row 1: Given pH = 3.5
[tex]\[ \text{pH} + \text{pOH} = 14 \Rightarrow \text{pOH} = 14 - \text{pH} = 14 - 3.5 = 10.5 \][/tex]
Next, we'll use [tex]\([\text{H}^+] = 10^{-\text{pH}}\)[/tex].
[tex]\[ [\text{H}^+] = 10^{-3.5} \][/tex]
This can be approximated but is commonly used:
[tex]\[ [\text{H}^+] \approx 3.16 \times 10^{-4} \][/tex]
Lastly, using the relationship [tex]\([\text{H}^+][\text{OH}^-] = 1 \times 10^{-14}\)[/tex]:
[tex]\[ [\text{OH}^-] = \frac{1 \times 10^{-14}}{[\text{H}^+]} = \frac{1 \times 10^{-14}}{3.16 \times 10^{-4}} \approx 3.16 \times 10^{-11} \][/tex]
- pH = 3.5, pOH = 10.5, [\text{H}^+] = 3.16 \times 10^{-4}, [\text{OH}^-] \approx 3.16 \times 10^{-11}
2. Row 2: Given pOH = 2.9
[tex]\[ \text{pH} + \text{pOH} = 14 \Rightarrow \text{pH} = 14 - \text{pOH} = 14 - 2.9 = 11.1 \][/tex]
To find [tex]\([\text{H}^+]\)[/tex]:
[tex]\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-11.1} \][/tex]
This can be approximated as:
[tex]\[ [\text{H}^+] \approx 7.94 \times 10^{-12} \][/tex]
Lastly, using [tex]\([\text{H}^+][\text{OH}^-] = 1 \times 10^{-14}\)[/tex] again:
[tex]\[ [\text{OH}^-] = 10^{-2.9} \][/tex]
This value can be calculated easily as:
[tex]\[ [\text{OH}^-] \approx 1.26 \times 10^{-3} \][/tex]
- pH = 11.1, pOH = 2.9, [\text{H}^+] \approx 7.94 \times 10^{-12}, [\text{OH}^-] \approx 1.26 \times 10^{-3}
3. Row 3: Given [tex]\([\text{H}^+] = 4.3 \times 10^{-4}\)[/tex]
Using [tex]\(\text{pH} = -\log [\text{H}^+]\)[/tex]:
[tex]\[ \text{pH} = -\log (4.3 \times 10^{-4}) \approx 3.37 \][/tex]
Then, [tex]\(\text{pH} + \text{pOH} = 14\)[/tex]:
[tex]\[ \text{pOH} = 14 - 3.37 = 10.63 \][/tex]
Subsequently, [tex]\([\text{OH}^-]\)[/tex] can be calculated:
[tex]\[ [\text{OH}^-] = \frac{1 \times 10^{-14}}{4.3 \times 10^{-4}} \approx 2.33 \times 10^{-11} \][/tex]
- pH \approx 3.37, pOH \approx 10.63, [\text{H}^+] = 4.3 \times 10^{-4}, [\text{OH}^-] \approx 2.33 \times 10^{-11}
4. Row 4: Given [tex]\([\text{OH}^-] = 5.7 \times 10^{-8}\)[/tex]
Using [tex]\(\text{pOH} = -\log [\text{OH}^-]\)[/tex]:
[tex]\[ \text{pOH} = -\log (5.7 \times 10^{-8}) \approx 7.24 \][/tex]
Then, [tex]\(\text{pOH} + \text{pH} = 14\)[/tex]:
[tex]\[ \text{pH} = 14 - 7.24 = 6.76 \][/tex]
Subsequently, [tex]\([\text{H}^+]\)[/tex] can be calculated:
[tex]\[ [\text{H}^+] = \frac{1 \times 10^{-14}}{5.7 \times 10^{-8}} \approx 1.75 \times 10^{-7} \][/tex]
- pH \approx 6.76, pOH \approx 7.24, [\text{H}^+] \approx 1.75 \times 10^{-7}, [\text{OH}^-] = 5.7 \times 10^{-8}
Completed table:
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
pH & pOH & {[tex]$\left[ \text{H} ^{+}\right]$[/tex]} & {[tex]$\left[ \text{OH} ^{-1}\right.$[/tex]} \\
\hline
3.5 & 10.5 & [tex]$3.16 \times 10^{-4}$[/tex] & [tex]$3.16 \times 10^{-11}$[/tex] \\
\hline
11.1 & 2.9 & [tex]$7.94 \times 10^{-12}$[/tex] & [tex]$1.26 \times 10^{-3}$[/tex] \\
\hline
3.37 & 10.63 & [tex]$4.3 \times 10^{-4}$[/tex] & [tex]$2.33 \times 10^{-11}$[/tex] \\
\hline
6.76 & 7.24 & [tex]$1.75 \times 10^{-7}$[/tex] & [tex]$5.7 \times 10^{-8}$[/tex] \\
\hline
\end{tabular}
\end{center}
