91Ó°ÊÓ

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 $1.0×{10}^{-7}M$at ${25}^{0}C$is

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

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

The given is baking soda,$1.0×{10}^{-6}M$.

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

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

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

The given is blood,$2.5×{10}^{-7}M$.

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

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

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

The given is milk,$4.0×{10}^{-7}M$

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

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

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

The given is bleach, $2.1×{10}^{-3}M$

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

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

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