91Ó°ÊÓ

The given is coffee,$1.0×{10}^{-5}M$.

We have to find therole="math" localid="1652596113018" $\left[O{H}^{-}\right]$of the solution.

Fill in the $K_w$for water.

$K_w=\left[H_3{O}^{+}\right]×\left[O{H}^{-}\right]=1.0×{10}^{-14}$

If you rearrange the $K_w$for $O{H}^{-}$, you get

$\frac{K_w}{\left[H_3{O}^{+}\right]}=\frac{\left[\cancel{H_3{O}^{+}}\right]×\left[O{H}^{-}\right]}{\cancel{[H_3{O}^{+}}]}=\frac{1.0×{10}^{-14}}{\left[H_3{O}^{+}\right]}$

$\left[O{H}^{-}\right]=\frac{1.0×{10}^{-14}}{\left[H_3{O}^{+}\right]}$

Calculate by substituting the known $\left[H_3{O}^{+}\right]$

$\left[O{H}^{-}\right]=\frac{1.0×{10}^{-14}}{[1.0×{10}^{-5}]}=1.0×{10}^{-9}M$

The given is soap, $1.0×{10}^{-8}M$.

We have to find the $\left[O{H}^{-}\right]$of the solution.

Fill in the $K_w$for water.

$K_w=\left[H_3{O}^{+}\right]×\left[O{H}^{-}\right]=1.0×{10}^{-14}$

If you rearrange the $K_w$for $\left[O{H}^{-}\right]$, you get

$\frac{K_w}{\left[H_3{O}^{+}\right]}=\frac{\left[\cancel{H_3{O}^{+}}\right]×\left[O{H}^{-}\right]}{\cancel{[H_3{O}^{+}}]}=\frac{1.0×{10}^{-14}}{\left[H_3{O}^{+}\right]}$

$\left[O{H}^{-}\right]=\frac{1.0×{10}^{-14}}{\left[H_3{O}^{+}\right]}$

Calculate by substituting the known $\left[H_3{O}^{+}\right]$

$\left[O{H}^{-}\right]=\frac{1.0×{10}^{-14}}{[1.0×{10}^{-8}]}=1.0×{10}^{-6}M$

The given is cleanser, $5.0×{10}^{-10}M$

We have to find the$\left[O{H}^{-}\right]$of the solution.

Fill in the $K_w$for water.

$K_w=\left[H_3{O}^{+}\right]×\left[O{H}^{-}\right]=1.0×{10}^{-14}$

Rearrange the $K_w$for the best results.

$\frac{K_w}{\left[H_3{O}^{+}\right]}=\frac{\left[\cancel{H_3{O}^{+}}\right]×\left[O{H}^{-}\right]}{\cancel{[H_3{O}^{+}}]}=\frac{1.0×{10}^{-14}}{\left[H_3{O}^{+}\right]}$

$\left[O{H}^{-}\right]=\frac{1.0×{10}^{-14}}{\left[H_3{O}^{+}\right]}$

Calculate by substituting the known $\left[H_3{O}^{+}\right]$

$\left[O{H}^{-}\right]=\frac{1.0×{10}^{-14}}{[1.0×{10}^{-10}]}=2.0×{10}^{-5}M$

The given is lemon juice, $2.5×{10}^{-2}M$

We have to find the$\left[O{H}^{-}\right]$of the solution

Fill in the $K_w$for water.

$K_w=\left[H_3{O}^{+}\right]×\left[O{H}^{-}\right]=1.0×{10}^{-14}$

Rearrange the $K_w$for the best results.

$\frac{K_w}{\left[H_3{O}^{+}\right]}=\frac{\left[\cancel{H_3{O}^{+}}\right]×\left[O{H}^{-}\right]}{\cancel{[H_3{O}^{+}}]}=\frac{1.0×{10}^{-14}}{\left[H_3{O}^{+}\right]}$

$\left[O{H}^{-}\right]=\frac{1.0×{10}^{-14}}{\left[H_3{O}^{+}\right]}$

Calculate by substituting the known $\left[H_3{O}^{+}\right]$

$\left[O{H}^{-}\right]=\frac{1.0×{10}^{-14}}{[1.0×{10}^{-2}]}=4.0×{10}^{-13}M$