# How do you convert the following p-functions to molar concentrations? pLi = -0.221; pMn = 0.0025; pNO_3 = 7.77

Apr 14, 2017

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

#### Explanation:

And so, by analogy, $p L i = - {\log}_{10} \left[L {i}^{+}\right]$

And thus if $p L i = - 0.221$, then ............................

$\left[L {i}^{+}\right] = {10}^{- \left(- 0.221\right)} = {10}^{0.221} = 1.66 \cdot m o l \cdot {L}^{-} 1$

$p M n = - {\log}_{10} \left[M {n}^{2 +}\right]$

And thus $\left[M {n}^{2 +}\right] = {10}^{- \left(- 0.0025\right)} = {10}^{0.0025} = 1.00 \cdot m o l \cdot {L}^{-} 1$

$p N {O}_{3}^{-} = - {\log}_{10} \left[N {O}_{3}^{-}\right]$

And thus $\left[N {O}_{3}^{-}\right] = {10}^{- 7.77} = 1.60 \times {10}^{-} 8 \cdot m o l \cdot {L}^{-} 1$

Back in the day, before the advent of cheap electronic calculators, which is not so long ago, students, engineers, and scientists, would routinely use tables of common logarithms to do these sorts of calculations. And we can still get them wrong these days.
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