Defining and Calculating pOH (HSC SSCE Chemistry): Flashcards

Hydroxide ion concentration

$\text{pOH} = -\log_{10}[\text{OH}^-]$

$[\text{OH}^-] = 10^{-\text{pOH}}$

$\text{pH} + \text{pOH} = 14$

$1.0 \times 10^{-14}$

1:1 or $[\text{NaOH}] = [\text{OH}^-]$

1:2 or $[\text{Ba(OH)}_2] = \frac{1}{2}[\text{OH}^-]$

1 unit

2 units

25°C or 298 K