Question

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

a. oven cleaner, $1.0\times {10}^{-12}M$

b. milk of magnesia, $1.0\times {10}^{-9}M$

c. aspirin, $6.0\times {10}^{-4}M$

d. pancreatic juice,$4.0\times {10}^{-9}M$

Short answer

Part a. $\left[O{H}^{-}\right]$is $1.0\times {10}^{-2}M$

Part b. $\left[O{H}^{-}\right]$is $1.0\times {10}^{-5}M$

Part c. $\left[O{H}^{-}\right]$is $1.67\times {10}^{-11}M$

Part d.$\left[O{H}^{-}\right]$is$2.5\times {10}^{-6}M$

Step-by-step solution

Calculation

Calculate $\left[H_3{O}^{+}\right]$and $\left[O{H}^{-}\right]$using the water dissociation expression.

The concentration of $\left[H_3{O}^{+}\right]$and $\left[O{H}^{-}\right]$in pure water is role="math" localid="1652599472836" $1.0\times {10}^{-7}M$at ${25}^{0}C$is

$K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]......\left(1\right)$

$$\begin{gathered} K_w=[1.0\times {10}^{-7}][1.0\times {10}^{-7}] \\ {K}_w=1.0\times {10}^{-14} \end{gathered}$$

Introduction (part a)

The given is oven cleaner,$1.0\times {10}^{-12}M$.

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

Explanation (part a)

The concentration of $\left[H_3{O}^{+}\right]$(oven cleaner)given is $1.0\times {10}^{-12}M$. Using equation $\left(1\right)$, calculate the $\left[O{H}^{-}\right]$concentration.

$$\begin{gathered} K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right] \\ 1.0\times {10}^{-14}=[1.0\times {10}^{-12}]\left[O{H}^{-}\right] \\ \left[O{H}^{-}\right]=\frac{1.0\times {10}^{-14}{M}^2}{1.0\times {10}^{-12}M} \\ =1.0\times {10}^{-2}M \end{gathered}$$

Introduction (part b)

The given is milk of magnesia, $1.0\times {10}^{-9}M$

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

Explanation (part b)

The concentration of $\left[H_3{O}^{+}\right]$(milk of magnesia) given is $1.0\times {10}^{-9}M$. Using equation$\left(1\right)$, calculate the $\left[O{H}^{-}\right]$concentration.

$$\begin{gathered} K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right] \\ 1.0\times {10}^{-14}=[1.0\times {10}^{-9}]\left[O{H}^{-}\right] \\ \left[O{H}^{-}\right]=\frac{1.0\times {10}^{-14}{M}^2}{1.0\times {10}^{-9}M} \\ =1.0\times {10}^{-5}M \end{gathered}$$

Introduction (part c)

The given is aspirin, $6.0\times {10}^{-4}M$

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

Explanation (part c)

The concentration of $\left[H_3{O}^{+}\right]$(aspirin) given is $6.0\times {10}^{-4}M$. Using equation $\left(1\right)$, calculate $\left[O{H}^{-}\right]$concentration.

$$\begin{gathered} K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right] \\ 1.0\times {10}^{-14}=[6.0\times {10}^{-4}]\left[O{H}^{-}\right] \\ \left[O{H}^{-}\right]=\frac{1.0\times {10}^{-14}{M}^2}{6.0\times {10}^{-4}M} \\ =1.6\times {10}^{-11}M \end{gathered}$$

Introduction (part d)

The given is pancreatic juice, $4.0\times {10}^{-9}M$

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

Explanation(part d)

The concentration of $\left[H_3{O}^{+}\right]$(pancreatic juice) given is $4.0\times {10}^{-9}M$. Using equation $\left(1\right)$, calculate $\left[O{H}^{-}\right]$concentration.

$$\begin{gathered} K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right] \\ 1.0\times {10}^{-14}=[4.0\times {10}^{-9}]\left[O{H}^{-}\right] \\ \left[O{H}^{-}\right]=\frac{1.0\times {10}^{-14}{M}^2}{4.0\times {10}^{-9}M} \\ =2.5\times {10}^{-6}M \end{gathered}$$