In Exercises $$ 13-16 $$, use the following information. Negative ions, designated by the notation $$ \left[\mathrm{OH}^{-}\right] $$, are always present in any acid or base. The concentration, $$ \left[\mathrm{OH}^{-}\right] $$, of these ions is related to $$ \left[\mathrm{H}^{+}\right] $$by the equation $$ \left[\mathrm{OH}^{-}\right] \cdot\left[\mathrm{H}^{+}\right]=10^{-14} $$ moles per liter. Find the concentrations of $$ \left[\mathrm{OH}^{-}\right] $$and $$ \left[\mathrm{H}^{+}\right] $$in moles per liter for the substances with the following pH values. 13. $$ \mathrm{pH}=6 $$ 14. $$ \mathrm{pH}=8 $$ 15. $$ \mathrm{pH}=9.5 $$

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To find the concentrations of $$ \left[\mathrm{OH}^{-}\right] $$ and $$ \left[\mathrm{H}^{+}\right] $$ in moles per liter for the substances with the given pH values, we can use the relationship between pH and the concentration of hydrogen ions ($$ \left[\mathrm{H}^{+}\right] $$):

$$ \mathrm{pH} = -\log_{10}\left[\mathrm{H}^{+}\right] $$

From this equation, we can solve for $$ \left[\mathrm{H}^{+}\right] $$:

$$ \left[\mathrm{H}^{+}\right] = 10^{-\mathrm{pH}} $$

Once we have $$ \left[\mathrm{H}^{+}\right] $$, we can use the equation $$ \left[\mathrm{OH}^{-}\right] \cdot\left[\mathrm{H}^{+}\right] = 10^{-14} $$ to find $$ \left[\mathrm{OH}^{-}\right] $$.

Let's calculate the concentrations for each pH value:

13. $$ \mathrm{pH} = 6 $$

$$ \left[\mathrm{H}^{+}\right] = 10^{-6} \text{ moles per liter} $$

Now, using the relationship between $$ \left[\mathrm{OH}^{-}\right] $$ and $$ \left[\mathrm{H}^{+}\right] $$:

$$ \left[\mathrm{OH}^{-}\right] = \frac{10^{-14}}{\left[\mathrm{H}^{+}\right]} = \frac{10^{-14}}{10^{-6}} = 10^{-8} \text{ moles per liter} $$

14. $$ \mathrm{pH} = 8 $$

$$ \left[\mathrm{H}^{+}\right] = 10^{-8} \text{ moles per liter} $$

$$ \left[\mathrm{OH}^{-}\right] = \frac{10^{-14}}{\left[\mathrm{H}^{+}\right]} = \frac{10^{-14}}{10^{-8}} = 10^{-6} \text{ moles per liter} $$

15. $$ \mathrm{pH} = 9.5 $$

$$ \left[\mathrm{H}^{+}\right] = 10^{-9.5} \text{ moles per liter} $$

$$ \left[\mathrm{OH}^{-}\right] = \frac{10^{-14}}{\left[\mathrm{H}^{+}\right]} = \frac{10^{-14}}{10^{-9.5}} = 10^{-4.5} \text{ moles per liter} $$

So, the concentrations of $$ \left[\mathrm{OH}^{-}\right] $$ and $$ \left[\mathrm{H}^{+}\right] $$ in moles per liter for the substances with the given pH values are:

13. $$ \left[\mathrm{OH}^{-}\right] = 10^{-8} \text{ M}, \left[\mathrm{H}^{+}\right] = 10^{-6} \text{ M} $$
14. $$ \left[\mathrm{OH}^{-}\right] = 10^{-6} \text{ M}, \left[\mathrm{H}^{+}\right] = 10^{-8} \text{ M} $$
15. $$ \left[\mathrm{OH}^{-}\right] = 10^{-4.5} \text{ M}, \left[\mathrm{H}^{+}\right] = 10^{-9.5} \text{ M} $$
13. $$ \left[\mathrm{OH}^{-}\right] = 10^{-8} \text{ M}, \left[\mathrm{H}^{+}\right] = 10^{-6} \text{ M} $$ 14. $$ \left[\mathrm{OH}^{-}\right] = 10^{-6} \text{ M}, \left[\mathrm{H}^{+}\right] = 10^{-8} \text{ M} $$ 15. $$ \left[\mathrm{OH}^{-}\right] = 10^{-4.5} \text{ M}, \left[\mathrm{H}^{+}\right] = 10^{-9.5} \text{ M} $$

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