# What is the hydrogen ion concentration in a urine specimen that registers a pH of 4 on a strip of pH paper?

Jul 17, 2017

Well, $\left[{H}_{3} {O}^{+}\right] = 1 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1. \ldots \ldots$

#### Explanation:

And I didn't even need a calculator.........

By definition, $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$.

And if $p H = 4$, then...........

$\left[{H}_{3} {O}^{+}\right] = {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1 = 0.0001 \cdot m o l \cdot {L}^{-} 1.$

And we know that hydroxide/hydronium ions obey the following equilibrium in aqueous solution.......

${H}_{3} {O}^{+} + H {O}^{-} \rightarrow 2 {H}_{2} O$; ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$. Of course, this value HAS to be measured, and here ${K}_{w}$ is measured under standard conditions of $298 \cdot K$ and (almost) $1 \cdot a t m$.

Why do we use such an absurd definition? Well, back in the day, approx. 30-40 years ago BEFORE the proliferation of hand-held electronic calculators, students and engineers would routinely use log tables for lengthy calculation involving multiplication and division. because the product $a \times b = {10}^{{\log}_{10} a + {\log}_{10} b}$, and it was easier to do additions than multiplications, even if we had to take antilogs to get the final product.

These days with a simple electronic calculator (and I bought one for a quid in a discounter's shop a couple of weeks ago) we have access to a computational power that would have astonished both Newton and Gauss (they say that Gauss in particular, a mathematical prodigy, had memorized the log tables so that he could do his calculations).