Q. 10.93 - An Introduction to General, Organic, and Biological Chemistry

Question

Solution$A$ has a pH of$4.0$, and solution B has a $\mathrm{pH}$ of$6.0$. (10.6)

a. Which solution is mone acidic?

b. What is the $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$in each?

c. What is the $\left[{\mathrm{OH}}^{-}\right]$in each?

Step-by-Step Solution

Answer

(Part a). Therefore $\mathrm{pH}$ of solution A will be more acidic

(Part b) $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.0\times {10}^{-8} \mathrm{M}$

(Part c) $\left[{\mathrm{OH}}^{-}\right]=1.0\times {10}^{-8} \mathrm{M}$

Step 1: Given information

That are given sentences are :(a). 

The $\mathrm{pH}$ of solution $\mathrm{A}$ is less than$\mathrm{pH}$of solution $\mathrm{B}$,

Step 2: Explanation (a)

(a) Since the $\mathrm{pH}$ of solution $\mathrm{A}$ is less than $\mathrm{pH}$ of solution $\mathrm{B}$, therefore $\mathrm{pH}$of solution $\mathrm{A}$ will be more acidic.

Step 3: Given information Part b

Find$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$.

Step 4: Explanation (b)

(b)

$pH=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

$\therefore \log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=-pH$

$=-4.0(p H=4.0 given for solution )$

$\therefore \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=$anti$\log (-4.0)$

$= anti \log (4.0)$

$= 1\times {10}^{-4}\mathrm{M}$

Step 5: Given information part c

Find concentration of hydroxide ion.

Step 6: Explanation (c)

For solution A

(c) We know that,

$$\begin{gathered} \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=K_e=1.0\times {10}^{-14} \\ \left[{\mathrm{OH}}^{-}\right]=\frac{K_e}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}=\frac{1.0\times {10}^{-14}}{1\times {10}^{-4}}=1\times {10}^{-10} \end{gathered}$$