Question

Question #44429

Answer 1

`[OH^-]=1.74xx10^-6M`

Explanation:

First, we use this relationship between pH and pOH:
`"pH"+"pOH"=14`

We know that `"pH"=8.24`, so
`"pOH"=14-"pH"=14-8.24=5.76`

To determine the `[OH^-]`, we use the following relationship between `"pOH"` and `[OH^-]`:
`-log[OH^-]="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 `log100=2`. In general, `logx=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 `"pOH"=5.76`, so using the formula above, we have:
`-log[OH^-]=5.76`

Dividing by `-1` to transfer the negative sign, we have:
`log[OH^-]=-5.76`

Finally, we raise both sides to the power of ten to cancel the logarithm, leaving us with:
`[OH^-]=10^-5.76=1.74xx10^-6M`
as our concentration.