Question

Calculate the hydroxide ion concentration [oh-] for human urine (ph = 6.2). notice this is about hydroxide.

Answer 1

[OH⁻] = 1.6 × 10⁻⁸ mol / dm³

Explanation

By definition, $[\text{H}^{+}] = 10^{-\text{pH}}$, where $[\text{H}^{+}]$ is the concentration of proton in the solution.

pH = 6.2 for this solution. As a result, $[\text{H}^{+} = 10^{-6.2} = 6.31 \times 10^{-7} \; \text{mol} \cdot \text{dm}^{-3}$.

$[\text{H}^{+}] \cdot [\text{OH}^{-}] = \text{K}_w$, where $[\text{OH}^{-}]$ the concentration of hydroxide ions and $\text{K}_w$ is the dissociation constant of water.

$\text{K}_w = 10^{-14} \; \text{mol}\cdot \text{dm}^{-3}$ at 0.10 MPa and 25 °C. As a result, $[\text{OH}^{-}] = \text{K}_w / [\text{H}^{+}] = 10^{14} / (6.31 \times 10^{-7}) = 1.6 \times 10^{-8} \; \text{mol}\cdot \text{dm}^{-3}$.