Q.10.111 - An Introduction to General, Organic, and Biological Chemistry

Question

One of the most acidic lakes in the United States is Little Echo Pond in the Adirondacks in New York. Recently, this lake had a pH of 4.2, well below the recommended pH of 6.5.

a. What are the $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \text{and }\left[{\mathrm{OH}}^{-}\right]$of Little Echo Pond?

b. What is the $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \text{and }\left[{\mathrm{OH}}^{-}\right]$of a lake that has a pH of 6.5?

c. One way to raise the pH (and restore aquatic life) is to add limestone $CaC{O}_3$. How many grams of $CaC{O}_3$ are needed to neutralize $1.0 kL$ of the acidic water from Little Echo Pond if the acid is sulfuric acid?

${\mathrm{H}}_2{\mathrm{SO}}_4 \left(aq\right)+{\mathrm{CaCO}}_3 \left(s\right)⟶{\mathrm{CO}}_2 \left(g\right)+{\mathrm{H}}_2\mathrm{O} \left(l\right)+{\mathrm{CaSO}}_4 \left(aq\right)$

Step-by-Step Solution

Answer

Part (a). $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=6\times {10}^{-5} \mathrm{M}\text{ and }\left[{\mathrm{OH}}^{-}\right]=2\times {10}^{-10} \mathrm{M}$

Part (b). $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=3\times {10}^{-7 }\mathrm{M}\text{ and }\left[{\mathrm{OH}}^{-}\right]=3\times {10}^{-8 }\mathrm{M}$

Part (c). $3 g$

Step 1: Given Information (Part a)

pH is a significant amount that mirrors the substance states of an answer. The pH has some control over the accessibility of supplements, organic capacities, microbial action, and the way of behaving of synthetic compounds.

Step 2: Explanation Part (a)

We know, pH= $4.2$

Finding $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$,

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-\mathrm{pH}}$

$$\begin{gathered} ={10}^{-4.2} \\ =6\times {10}^{-5 }M \end{gathered}$$

Using,

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \left[{\mathrm{OH}}^{-}\right]=1.0\times {10}^{-14}$

$$\begin{gathered} \left[{\mathrm{OH}}^{-}\right]=\frac{{\displaystyle 1.0\times {10}^{-14}}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}=\frac{1\times {10}^{-14}}{6\times {10}^{-5}} \\ =2\times {10}^{-10} M \end{gathered}$$

Step 3: Explanation Part (b)

We know, pH= $6.5$

Finding$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$,

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-\mathrm{pH}}$$={10}^{-6.5}=3\times {10}^{-7} \mathrm{M}$

Using,

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \left[{\mathrm{OH}}^{-}\right]=1.0\times {10}^{-14}$

$\left[{\mathrm{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{3\times {10}^{-7}}=3\times {10}^{-8} M$

Step 4: Explanation Part (c)

The balanced reaction is,

${\mathrm{H}}_2{\mathrm{SO}}_4 \left(aq\right) + {\mathrm{CaCO}}_3 \left(s\right)\to {\mathrm{CaSO}}_4 \left(aq\right)+{\mathrm{CO}}_2 \left(g\right)+{\mathrm{H}}_2\mathrm{O} \left(l\right)$

The ratio of the no. of moles is$1:2$, hence,

$6\times {10}^{-5} \mathrm{M} {\mathrm{H}}_3{\mathrm{O}}^{+}\times \frac{1 {\mathrm{M} \mathrm{H}}_2{\mathrm{SO}}_4}{2 {\mathrm{M} \mathrm{H}}_3{\mathrm{O}}^{+}}=3\times {10}^{-5} \mathrm{M} {\mathrm{H}}_2{\mathrm{SO}}_4$

From the reaction $1 \mathrm{mol} {\mathrm{H}}_2{\mathrm{SO}}_4=1 \mathrm{mol} {\mathrm{CaCO}}_3$

Calculating the molar mass of $CaC{O}_3$,

$1 \mathrm{mol} {\mathrm{CaCO}}_3=100.09 \mathrm{g} {\mathrm{CaCO}}_3$

The required amount of $CaC{O}_3$,

$1 \mathrm{kL} {\mathrm{H}}_2{\mathrm{SO}}_4\times \frac{1000 {\mathrm{L} \mathrm{H}}_2{\mathrm{SO}}_4}{1 \mathrm{kL} {\mathrm{H}}_2{\mathrm{SO}}_4}\times \frac{3\times {10}^{-5} \mathrm{mol} {\mathrm{H}}_2{\mathrm{SO}}_4}{1 {\mathrm{L} \mathrm{H}}_2{\mathrm{SO}}_4}\times \frac{1 \mathrm{mol} {\mathrm{CaCO}}_3}{1 \mathrm{mol} {\mathrm{H}}_2{\mathrm{SO}}_4}\times \frac{100.09 \mathrm{g} {\mathrm{CaCO}}_3}{1 \mathrm{mol} {\mathrm{CaCO}}_3}$

$=3 g$