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

a. stomach acid, $2.5\times {10}^{-13}M$

b. urine, $2.0\times {10}^{-9}M$

c. orange juice, $5.0\times {10}^{-11}M$

d. bile,$2.5\times {10}^{-6}M$

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 $1.0\times {10}^{-7}M$at ${25}^{0}C$is

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

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

The given is stomach acid, $2.5\times {10}^{-13}M$

We have to find the$\left[H_3{O}^{+}\right]$of the solution.

The concentration of $\left[O{H}^{-}\right]$(stomach acid) given is $2.5\times {10}^{-13}M$. Using equation $\left(1\right)$, calculate $\left[H_3{O}^{+}\right]$concentration.

localid="1652608368007" $$\begin{gathered} K_w=\left[O{H}^{-}\right]\left[H_3{O}^{+}\right] \\ 1.0\times {10}^{-14}=[2.5\times {10}^{-13}]\left[H_3{O}^{+}\right] \\ \left[H_3{O}^{+}\right]=\frac{[1.0\times {10}^{-14}]M^2}{[2.5\times {10}^{-13}]M} \\ =4.0\times {10}^{-2}M \end{gathered}$$

The given is urine,$2.0\times {10}^{-9}M$

We have to find the$\left[H_3{O}^{+}\right]$of the solution.

The concentration of $\left[O{H}^{-}\right]$(urine) given is $2.0\times {10}^{-9}M$. Using equation $\left(1\right)$, calculate $\left[H_3{O}^{+}\right]$concentration.

localid="1652608390913" $$\begin{gathered} K_w=\left[O{H}^{-}\right]\left[H_3{O}^{+}\right] \\ 1.0\times {10}^{-14}=[2.0\times {10}^{-9}]\left[H_3{O}^{+}\right] \\ \left[H_3{O}^{+}\right]=\frac{[1.0\times {10}^{-14}]M^2}{[2.0\times {10}^{-9}]M} \\ =5.0\times {10}^{-6}M \end{gathered}$$

The given is orange juice,$5.0\times {10}^{-11}M$

We have to find the $\left[H_3{O}^{+}\right]$of the solution.

The concentration of $\left[O{H}^{-}\right]$(orange juice) given is $5.0\times {10}^{-11}M$. Using equation $\left(1\right)$, calculate $\left[H_3{O}^{+}\right]$concentration.

localid="1652608417361" $$\begin{gathered} K_w=\left[O{H}^{-}\right]\left[H_3{O}^{+}\right] \\ 1.0\times {10}^{-14}=[5.0\times {10}^{-11}]\left[H_3{O}^{+}\right] \\ \left[H_3{O}^{+}\right]=\frac{[1.0\times {10}^{-14}]M^2}{[5.0\times {10}^{-11}]M} \\ =2.0\times {10}^{-4}M \end{gathered}$$

The given is bile,$2.5\times {10}^{-6}M$

We have to find the$\left[H_3{O}^{+}\right]$of the solution.

The concentration of $\left[O{H}^{-}\right]$(bile) given is $2.5\times {10}^{-6}M$. Using equation $\left(1\right)$, calculate $\left[H_3{O}^{+}\right]$concentration.

localid="1652608442072" $$\begin{gathered} K_w=\left[O{H}^{-}\right]\left[H_3{O}^{+}\right] \\ 1.0\times {10}^{-14}=[2.5\times {10}^{-6}]\left[H_3{O}^{+}\right] \\ \left[H_3{O}^{+}\right]=\frac{[1.0\times {10}^{-14}]M^2}{[2.5\times {10}^{-6}]M} \\ =4.0\times {10}^{-9}M \end{gathered}$$