Question

(a)Use data from Tables 15.2 and 16.2 to calculate the solubility of${\mathrm{CaCO}}_3$in water at pH 7.

(b)Will the solubility increase or decrease if the pH is lowered and the water becomes more acidic?

(c)Calculate the maximum amount of limestone (primarily calcium carbonate) that could dissolve per year in a river at pH 7 with an average flow rate of$1.0\times {10}^6{\mathrm{m}}^3/\mathrm{h}$.

Short answer

a) The solubility of ${\mathrm{CaCO}}_3$is$0.0029$

b) If the pH is lowered and the water becomes more acidic the solubility is increases.

c) $0.052\times {10}^{13}{\mathrm{m}}^3$

Step-by-step solution

a)

From Tables 15.2 and 16.2 the solubility product is$\left[{\mathrm{Ca}}^{2+}\right]\left[{\mathrm{CO}}_3^{2-}\right]=8.7\times {10}^{-6}$

The reaction of ${\mathrm{CaCO}}_3$in water, and if the concentrations are taken x then we can calculate the solubility of the respective substance,

Hence the solubility of ${\mathrm{CaCO}}_3$is

$$\begin{gathered} 8.7\times {10}^{-6}={\mathrm{x}}^2 \\ \Rightarrow \mathrm{x}=0.0029 \end{gathered}$$

The solubility of ${\mathrm{CaCO}}_3$is$0.0029$

b)

If the pH is lowered and the water becomes more acidic the solubility is increases because the ionic compound containing basic anions it is increased and for the anions it basically unaffected.

c)

To calculate the volume of limestone present in river we can conclude the formula of average rate as,

$$\begin{gathered} \mathrm{F}=\frac{\mathrm{V}}{\mathrm{T}} \\ \Rightarrow 1.0\times {10}^6{\mathrm{m}}^3/\mathrm{h}=\frac{\mathrm{V}}{1\mathrm{Year}} \\ \Rightarrow 1.0\times {10}^6\frac{{\mathrm{m}}^3}{60\sec }\times 3.1536\times {10}^7\sec =\mathrm{v} \\ \Rightarrow \mathrm{V}=0.052\times {10}^{13}{\mathrm{m}}^3 \end{gathered}$$