pOH, pKw, pKa and pKb

Understanding pOH, pK_w, pK_a, and pK_b is essential for grasping the intricacies of acid-base chemistry, as they define the relationships between concentrations of hydrogen ions and their corresponding bases in aqueous solutions.

Extending on the concept of pH, we can define the negative logarithms of [OH^–], K_w, K_aand K_bas follows:

$pOH=-log[OH^-]\; \; \; \; \; \; \; \; (3)$

$pK_w=-logK_w\; \; \; \; \; \; \; \; (4)$

$pK_a=-logK_a\; \; \; \; \; \; \; \; (5)$

$pK_b=-logK_b\; \; \; \; \; \; \; \; (6)$

Since K_w = [H⁺][OH^–],

$-logK_w=-log[H^+]-log[OH^-]\; \; \; \; \; \; \; \; (7)$

Substitute eq1, eq3 and eq4 in eq7

$pK_w=pH+pOH$

At 25⁰C, K_w = 10^-14. So, pK_w= 14 and

$pH+pOH=14$

Question

The concentration of hydroxide ions in the urine of a patient with kidney stones is 1.698 x 10^-10 M. What is the pH of the patient’s urine?

Answer

$pOH=-log(1.698\times 10^{-10})$

$pH=14-[-log(1.698\times 10^{-10})]=4.23$

pOH, pKw, pKa and pKb

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