# What is the pH of a solution with [H+] = 2.3 times 10^-3?

##### 2 Answers
Jun 25, 2017

pH is defined as 'the logarithm (to base 10) of the [H+] with the sign changed'. $L o {g}_{10} \left(2.3 \times {10}^{-} 3\right) = - 2.64$. Change the sign to positive and the pH is 2.64.

Jun 25, 2017

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = - {\log}_{10} \left(2.3 \times {10}^{-} 3\right) = 2.64$

#### Explanation:

The former would be all you needed to write in an exam. But as to background, we KNOW from classic experiments that water undergoes autoprotolysis, i.e. self-ionization.

We could represent this reaction by $\left(i\right)$:

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

OR by $\left(i i\right)$:

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

Note that $\left(i\right)$ and $\left(i i\right)$ ARE EQUIVALENT REPRESENTATIONS, and it really is a matter of preference which equation you decide to use. As far as anyone knows, the actual acidium ion in solution is
${H}_{5} {O}_{2}^{+}$ or ${H}_{7} {O}_{3}^{+}$, i.e. a cluster of 2 or 3 or 4 (or more) water molecules with an EXTRA ${H}^{+}$ tacked on. We can use ${H}^{+}$, $\text{protium ion}$ (as is used here) or ${H}_{3} {O}^{+}$, $\text{hydronium ion}$ equivalently to represent this species.

The equilibrium constant for the reaction, under standard conditions, is..........K_w=[H_3O^+][""^(-)OH]=10^-14.

And to make the arithmetic a bit easier we can use the $p H$ function, where $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$. And thus in aqueous solution under the given standard conditions, $p H + p O H = 14$.

And since we are directly given $\left[{H}^{+}\right]$ or $\left[{H}_{3} {O}^{+}\right]$.....

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = - {\log}_{10} \left(2.3 \times {10}^{-} 3\right) = 2.64$

How do you think $p H$ and $p O H$ would evolve under non-standard conditions, i.e. at a temperature of say $373 \cdot K$? Would ${K}_{w}$ decrease, remain constant, increase? Remember that the reaction as written is A BOND BREAKING reaction. How would this inform your choice?