Q. 10.33 - An Introduction to General, Organic, and Biological Chemistry

Question

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, $2.5\times {10}^{-6}M$

Step-by-Step Solution

Answer

Part a. $\left[H_3{O}^{+}\right]$ is $4.0\times {10}^{-2}M$

Part b. $\left[H_3{O}^{+}\right]$ is $5.0\times {10}^{-6}M$

Part c. $\left[H_3{O}^{+}\right]$ is $2.0\times {10}^{-4}M$

Part d. $\left[H_3{O}^{+}\right]$ is $4.0\times {10}^{-9}M$

Step 1: 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 $1.0\times {10}^{-7}M$ at ${25}^{0}C$ is

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

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

Step 2: Introduction (part a)

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

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

Step 3: Explanation (part a)

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.

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

Step 4: Introduction (part b)

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

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

Step 5: Explanation (part b)

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.

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

Step 6: Introduction (part c)

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

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

Step 7: Explanation(Part c)

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.

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

Step 8: Introduction (part d)

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

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

Step 9:

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.

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