Find [OH-] of 0.325 M (NH4)2SO4 and 0.490 M NH3
To find the hydroxide ion concentration, [OH-], we need to consider the dissociation of the compounds (NH4)2SO4 and NH3.
1. (NH4)2SO4 dissociates into 2 NH4+ ions and 1 SO4 2- ion.
So, the concentration of NH4+ ions is twice the concentration of (NH4)2SO4, which is 0.325 M * 2 = 0.65 M.
2. NH3 is a weak base and can accept a proton to form NH4+ ion and hydroxide ion (OH-). For every 1 mole of NH3 that reacts with water, it forms 1 mole of NH4+ and 1 mole of OH-.
Now, we can calculate the concentration of OH-.
Since the NH4+ ion concentration (0.65 M) comes from (NH4)2SO4, we need to subtract this concentration from the NH3 concentration to determine the leftover NH3 that reacts with water to form OH-.
0.49 M - 0.65 M = -0.16 M
However, this calculated concentration is negative, which is not possible for the hydroxide ion concentration. It indicates that there is an insufficient amount of NH3 to fully react with the provided concentration of (NH4)2SO4.
Therefore, the hydroxide ion concentration in this solution is 0 M.