Question

Calculate the pH and pOH of aqueous solutions with the following concentrations:[H+] = 3.6 x 10-10M

Answer 1

Answer: the pH of the given solution is 9.4 and the pOH is 4.6

Explanation:

The question requires us to calculate the pH and pOH values of an aqueous solution that presents a concentration of H+ ions of 3.6 x 10^-10 M.

By definition, the pH of a solution can be calculated applying the negative logarithm of the H+ ions in solution, as it follows:

$pH=-\log_{10}[H^+\rbrack$

(where [H+] corresponds to the concentration of H+ ions in solution)

Thus, we can determine the pH of the solution provided by the question as:

$\begin{gathered} pH=-\operatorname{\log}_{10}[H^+\rbrack \\ \\ pH=-\log_{10}(3.6\times10^{-10})=9.4 \end{gathered}$

Therefore, the pH of the solution is 9.4.

Similarly, the pOH of a solution can be calculated as the negative logarithm of the OH- ions concentration in solution. Since the question does not provide the concentration of OH- ions, we can apply a different equation, which comes from the self-ionization of water.

The equilibrium constant for the self-ionization of water (a reaction where water dissociates into H+ and OH- ions) is 1.0 x 10^-14. Based on this, we can write that:

$14=pH+pOH$

And, with the calculated value of pH (pH = 9.4), we can determine the pOH value of this solution:

$14=9.4+pOH\rightarrow pOH=14-9.4\rightarrow pOH=4.6$

Therefore, the pOH of the solution is 4.6.

In summary, the pH of the given solution is 9.4 and the pOH is 4.6.