pOH, pKw, pKa and pKb

Extending on the concept of pH, we can define the negative logarithm of [OH^–], K_w, K_aand K_brespectively as 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

$pH+pOH=14$

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?

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

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