Concept:The relationship between pH and pOH is derived directly from the auto-ionization constant of water (\( K_w \)) at standard room temperature.
Formula:$$ K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \text{ at } 25^{\circ}\text{C} $$
Solution:- Take the negative base-10 logarithm of both sides of the \( K_w \) equation.
- \( -\log(K_w) = -\log([\text{H}^+][\text{OH}^-]) \)
- \( \text{pK}_w = -\log[\text{H}^+] + -\log[\text{OH}^-] \)
- \( \text{pK}_w = \text{pH} + \text{pOH} \).
- Since \( K_w = 10^{-14} \), taking the negative log gives exactly 14.
Why other options are incorrect:Options A and B confuse the absolute value of \( K_w \) (which is \( 10^{-14} \)) with the logarithmic scale (pH/pOH), which extracts just the positive exponent 14.
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