Question

Calculate the $\left[O{H}^{-}\right]$of each aqueous solution with the following $\left[H_3{O}^{+}\right]$:

a. coffee, $1.0\times {10}^{-5}M$

b. soap, $1.0\times {10}^{-8}M$

c. cleanser, $5.0\times {10}^{-10}M$

d. lemon juice,$2.5\times {10}^{-2}M$

Short answer

a. $\left[O{H}^{-}\right]$is $1.0\times {10}^{-9}M$

b. role="math" $\left[O{H}^{-}\right]$is$1.0\times {10}^{-6}M$

c. $\left[O{H}^{-}\right]$isrole="math" localid="1652595979772" $2.0\times {10}^{-5}M$

d.$\left[O{H}^{-}\right]$is$4.0\times {10}^{-13}M$

Step-by-step solution

Introduction (part a)

The given is coffee,$1.0\times {10}^{-5}M$.

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

Explanation (part a)

Fill in the $K_w$for water.

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

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

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

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

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

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

Introduction (part b)

The given is soap, $1.0\times {10}^{-8}M$.

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

Explanation (part b)

Fill in the $K_w$for water.

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

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

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

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

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

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

Introduction (part c)

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

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

Explanation (part c)

Fill in the $K_w$for water.

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

Rearrange the $K_w$for the best results.

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

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

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

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

Introduction (part d)

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

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

Explanation (part d)

Fill in the $K_w$for water.

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

Rearrange the $K_w$for the best results.

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

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

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

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