# Question #44429

Nov 7, 2016

$\left[O {H}^{-}\right] = 1.74 \times {10}^{-} 6 M$

#### Explanation:

First, we use this relationship between pH and pOH:
$\text{pH"+"pOH} = 14$

We know that $\text{pH} = 8.24$, so
$\text{pOH"=14-"pH} = 14 - 8.24 = 5.76$

To determine the $\left[O {H}^{-}\right]$, we use the following relationship between $\text{pOH}$ and $\left[O {H}^{-}\right]$:
$- \log \left[O {H}^{-}\right] = \text{pOH}$

Where "log" is the base ten logarithm. If you're not familiar with logs, they are essentially the opposite of exponents. For example, we know that ${10}^{2} = 100$; that means $\log 100 = 2$. In general, $\log x = y$ if ${10}^{y} = x$. If you want to learn more about logs, you can check this link.

Back to the problem. We know that $\text{pOH} = 5.76$, so using the formula above, we have:
$- \log \left[O {H}^{-}\right] = 5.76$

Dividing by $- 1$ to transfer the negative sign, we have:
$\log \left[O {H}^{-}\right] = - 5.76$

Finally, we raise both sides to the power of ten to cancel the logarithm, leaving us with:
$\left[O {H}^{-}\right] = {10}^{-} 5.76 = 1.74 \times {10}^{-} 6 M$
as our concentration.