To understand the relationship between pH and pOH, it's essential to know that they are measures of the acidity and basicity of an aqueous solution. The relationship between them is founded on the autoionization of water.
In aqueous solutions, water can dissociate into hydrogen ions (H⁺) and hydroxide ions (OH⁻):
[tex]\[ \text{H}_2\text{O} \leftrightarrow \text{H}^+ + \text{OH}^- \][/tex]
The product of the concentrations of these ions is a constant at a given temperature, specifically 25°C (298 K):
[tex]\[ [\text{H}^+] \cdot [\text{OH}^- ] = 1 \times 10^{-14} \][/tex]
To express the concentrations more conveniently, we use the pH and pOH scales:
[tex]\[ \text{pH} = -\log [\text{H}^+] \][/tex]
[tex]\[ \text{pOH} = -\log [\text{OH}^-] \][/tex]
Given the ion product of water:
[tex]\[ [\text{H}^+] \cdot [\text{OH}^- ] = 1 \times 10^{-14} \][/tex]
Taking the negative logarithm of both sides:
[tex]\[ -\log([\text{H}^+] \cdot [\text{OH}^- ]) = -\log(1 \times 10^{-14}) \][/tex]
Using properties of logarithms, this can be expanded to:
[tex]\[ -\log([\text{H}^+]) + (-\log([\text{OH}^-])) = -\log(1 \times 10^{-14}) \][/tex]
Hence:
[tex]\[ \text{pH} + \text{pOH} = 14 \][/tex]
So, if you solve for pOH, you get:
[tex]\[ \text{pOH} = 14 - \text{pH} \][/tex]
Therefore, the correct relationship between pH and pOH is:
[tex]\[ \text{pOH} = 14 - \text{pH} \][/tex]
Thus, the correct answer is:
B. [tex]\( \text{pOH} = 14 - \text{pH} \)[/tex]
