Question

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

a. baking soda, $1.0\times {10}^{-6}M$

b. blood, $2.5\times {10}^{-7}M$

c. milk, $4.0\times {10}^{-7}M$

d. bleach,$2.1\times {10}^{-3}M$

Short answer

Part a. $\left[H_3{O}^{+}\right]$is $1.0\times {10}^{-8}M$

Part b. $\left[H_3{O}^{+}\right]$is $4.0\times {10}^{-8}M$

Part c. $\left[H_3{O}^{+}\right]$is $2.5\times {10}^{-8}M$

Part d. $\left[H_3{O}^{+}\right]$is $4.76\times {10}^{-12}M$

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 $1.0\times {10}^{-7}M$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 baking soda,$1.0\times {10}^{-6}M$.

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

Explanation (part a)

The concentration of $\left[O{H}^{-}\right]$(baking soda) given is $1.0\times {10}^{-6}M$. Using equation $\left(1\right)$, calculate $\left[H_3{O}^{+}\right]$concentration.

role="math" localid="1652607935715" $$\begin{gathered} K_w=\left[O{H}^{-}\right]\left[H_3{O}^{+}\right] \\ 1.0\times {10}^{-14}=[1.0\times {10}^{-6}]\left[H_3{O}^{+}\right] \\ \left[H_3{O}^{+}\right]=\frac{[1.0\times {10}^{-14}]M^2}{[1.0\times {10}^{-6}]M} \\ =1.0\times {10}^{-8}M \end{gathered}$$

Introduction (part b)

The given is blood,$2.5\times {10}^{-7}M$.

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

Explanation (part b)

The concentration of $\left[O{H}^{-}\right]$(blood) given is $2.5\times {10}^{-7}M$. Using equation $\left(1\right)$, calculate $\left[H_3{O}^{+}\right]$concentration.

role="math" localid="1652607915433" $$\begin{gathered} K_w=\left[O{H}^{-}\right]\left[H_3{O}^{+}\right] \\ 1.0\times {10}^{-14}=[2.5\times {10}^{-7}]\left[H_3{O}^{+}\right] \\ \left[H_3{O}^{+}\right]=\frac{[1.0\times {10}^{-14}]M^2}{[2.5\times {10}^{-7}]M} \\ =4.0\times {10}^{-8}M \end{gathered}$$

Introduction(part c)

The given is milk,$4.0\times {10}^{-7}M$

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

Explanation (part c)

The concentration of $\left[O{H}^{-}\right]$(milk) given is $4.0\times {10}^{-7}M$. Using equation $\left(1\right)$, calculate $\left[H_3{O}^{+}\right]$concentration.

$$\begin{gathered} K_w=\left[O{H}^{-}\right]\left[H_3{O}^{+}\right] \\ 1.0\times {10}^{-14}=[4.0\times {10}^{-7}]\left[H_3{O}^{+}\right] \\ \left[H_3{O}^{+}\right]=\frac{[1.0\times {10}^{-14}]M^2}{[4.0\times {10}^{-7}]M} \\ =2.5\times {10}^{-8}M \end{gathered}$$

Introduction (part d)

The given is bleach, $2.1\times {10}^{-3}M$

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

Explanation (part d)

The concentration of $\left[O{H}^{-}\right]$(blood) given is $2.1\times {10}^{-3}M$. Using equation $\left(1\right)$, calculate $\left[H_3{O}^{+}\right]$concentration.

$$\begin{gathered} K_w=\left[O{H}^{-}\right]\left[H_3{O}^{+}\right] \\ 1.0\times {10}^{-14}=[2.1\times {10}^{-3}]\left[H_3{O}^{+}\right] \\ \left[H_3{O}^{+}\right]=\frac{[1.0\times {10}^{-14}]M^2}{[2.1\times {10}^{-3}]M} \\ =4.76\times {10}^{-12}M \end{gathered}$$