# Question 0059f

Jun 23, 2017

Here's how you can do that.

#### Explanation:

As you know, the $\text{pH}$ of a solution is calculated by taking the negative log base $10$ of the concentration of hydrogen ions, ${\text{H}}^{+}$, which you'll sometimes see referred to as hydronium ions, ${\text{H"_3"O}}^{+}$

"pH" = - log_10(["H"^(+)])

or, more simply

"pH" = - log(["H"^(+)])

In your case, the $\text{pH}$ of the solution is equal to $9.7$. Right from the start, the fact that you have

$\text{pH} > 7$

tells you that you're dealing with a basic solution, which, at room temperature, is a classification given to any solution that has

$\left[{\text{H}}^{+}\right] < {10}^{- 7}$ $\text{M}$

To find the actual concentration of hydrogen ions, rewrite the equation as

$\log \left(\left[{\text{H}}^{+}\right]\right) = - 9.7$

Now use both sides as exponents for $10$ to say that

${10}^{\log} \left(\left[{\text{H}}^{+}\right]\right) = {10}^{- 9.7}$

By definition, you have

${a}^{{\log}_{a} x} = x \text{ } \left(\forall\right) \textcolor{w h i t e}{.} a > 0 , x > 0 , a , x \in \mathbb{R}$

This means that

10^log(["H"^(+)]) = ["H"^(+)]#

which gets you

$\left[{\text{H}}^{+}\right] = {10}^{- 9.7}$

$\left[{\text{H}}^{+}\right] = 2.0 \cdot {10}^{- 10}$ $\text{M}$

As predicted, the concentration of hydrogen ions is $< {10}^{- 7}$ $\text{M}$, which is what you should expect to see for a basic solution.