Answer: for the given aqueous solution, the [OH-] value is 1.0 x 10^-11 M
Explanation:
The question requires us to calculate the concentration of OH- ions ([OH-]) in an aqueous solution, knowing that the concentration of H+ ions ([H+]) is 1.0 x 10^-3 M.
To solve this problem, we can consider the self-ionization of water and its correspondent ionization constant:
[tex]2H_2O_{(l)}\rightleftarrows H_3O_{(aq)}^++OH_{(aq)}^-\text{ K}_w=1.00\times10^{-14}[/tex]Note that the constant of equilibrium expression for the reaction above, Kw, can be written as:
[tex]K_w=[H_3O^+\rbrack\times[OH^-\rbrack[/tex]Also, note that H3O+ ions are equivalent to H+ ions.
Therefore, we can rearrange the equation above to calculate the concentration of OH- ions in an aqueous solution, knowing that the equilibrium constant for the self ionization of water is 1.00 x 10^-14 and that the concentration of H+ ions in the solution is 1.0 x 10^-3:
[tex]\begin{gathered} K_w=[H_3O^+\rbrack\times[OH^-\rbrack\rightarrow[OH^-\rbrack=\frac{K_w}{[H_3O^+\rbrack} \\ \\ [OH^-\rbrack=\frac{1.00\times10^{-14}}{1.0\times10^{-3}}=1.0\times10^{-11}M \end{gathered}[/tex]Therefore, for the given aqueous solution, the [OH-] value is 1.0 x 10^-11 M.