# How do you calculate the concentration of H^+ ions in a 0.27 M NaoH solution?

Mar 25, 2018

$\left[{H}_{3} {O}^{+}\right] = {10}^{- 13.43} \cdot m o l \cdot {L}^{-} 1 = 3.70 \times {10}^{-} 14 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We gots the equilibrium $2 {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

For which ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$ under standard conditions of temperature and pressure...and taking ${\log}_{10}$ of BOTH sides...

${\log}_{10} {K}_{w} = {\log}_{10} \left({10}^{-} 14\right) = {\underbrace{{\log}_{10} \left[{H}_{3} {O}^{+}\right]}}_{- p H} + {\underbrace{{\log}_{10} \left[H {O}^{-}\right]}}_{- p O H}$

Now logarithms were used in the old days before electronic calculators given that they allow the multiplication of large or small numbers fairly easily..

But by definition of the logarithmic function, ${\log}_{10} \left({10}^{-} 14\right) = - 14$...and so our defining relationship...

$p H + p O H = 14$

And we got $\left[H {O}^{-}\right] = 0.27 \cdot m o l \cdot {L}^{-} 1$...

$p O H = - {\log}_{10} \left(0.27\right) = - \left(- 0.569\right) = 0.569$

And $p H = 14 - p O H = 14 - 0.569 = 13.43$ as required....

And finally, we take antilogarithms...$\left[{H}_{3} {O}^{+}\right] = {10}^{- 13.43} = 3.70 \times {10}^{-} 14 \cdot m o l \cdot {L}^{-} 1$