Chapter 8, Problem 28P

Although we did not mention it in Sec. 8.2, Eq. $\left(8.10\right)$ is actually an expression of electroneutrality; that is, that positive and negative charges must balance. This can be seen more clearly by expressing it as

$\left[{\text{H}}^{+}\right]=\left[{\text{HCO}}_3^{-}\right]+2\left[{\text{CO}}_3^{2-}\right]+\left[{\text{OH}}^{-}\right]$

In other words, the positive charges must equal the negative charges. Thus, when you compute the pH of a natural water body such as a lake, you must also account for other ions that may be present. For the case where these ions originate from nonreactive salts, the net negative minus positive charges due to these ions are lumped together in a quantity called alkalinity, and the equation is reformulated as

$Alk+\left[{\text{H}}^{+}\right]=\left[{\text{HCO}}_3^{-}\right]+2\left[{\text{CO}}_3^{2-}\right]+\left[{\text{OH}}^{-}\right]\left(\text{P8}\text{.28}\right)$

where $Alk=\text{ alkalinity }\left(\text{eq}/\text{L}\right)$. For example, the alkalinity of Lake Superior is approximately $0.4\times {10}^{-3}\text{eq}/\text{L}$. Perform the same calculations as in Sec. 8.2 to compute the pH of Lake Superior in 2008. Assume that just like the raindrops, the lake is in equilibrium with atmospheric ${\text{CO}}_2$, but account for the alkalinity as in Eq. $\left(\text{P8}\text{.28}\right)$.