Ok here is another one!
Concentrated aqueous HClO4 has a concentration of 14.8 M. Calculate the concentrations of ALL ions present in a solution prepared by pipetting 5.00mL of concentrated HClO4 into a 1000.0-mL volumetric flask and filling to the mark. Also calculate the pH and pOH of the dilution.
Wow that was a mouth full! :)
(HClO4) = 14.8M x (5 mL/1000) = ??
HClO4 is a strong acid. It ionizes 100%; therefore, (H^+) = (ClO4^-) = (HClO4^-)
Convert H^+ to pH, then to pOH, then to OH^- and that's it. (unless I missed an ion somewhere).
To calculate the concentrations of all ions present in the solution, we need to understand that HClO4 dissociates in water into H+ and ClO4- ions.
First, let's calculate the moles of HClO4 that were transferred from the pipette into the 1000.0 mL volumetric flask:
Moles of HClO4 = concentration × volume in liters
Moles of HClO4 = (14.8 mol/L) × (5.00 mL / 1000.0 mL) = 0.074 mol
Since HClO4 is a strong acid, all of its moles will dissociate into H+ ions. Therefore, the concentration of H+ ions in the solution is equal to the concentration of HClO4:
Concentration of H+ ions = 14.8 M
Next, let's consider the ClO4- ions. Since HClO4 is a strong acid, its conjugate base, ClO4-, does not hydrolyze in water. Therefore, the concentration of ClO4- ions is also equal to the concentration of HClO4:
Concentration of ClO4- ions = 14.8 M
Now, let's calculate the pH of the dilution. Since the concentration of H+ ions is 14.8 M, we can use the pH formula:
pH = -log[H+]
pH = -log(14.8) ≈ -1.17
The pOH can be calculated by using the relationship:
pH + pOH = 14
pOH = 14 - pH
pOH = 14 - (-1.17) ≈ 15.17
Therefore, the concentrations of all ions present in the solution are:
[H+] = 14.8 M
[ClO4-] = 14.8 M
And the pH and pOH of the dilution are approximately pH = -1.17 and pOH = 15.17, respectively.