posted by Brittany on .
A 10g sample of aniline hyrdochloride is dissolved in 250 mL of solution.
C6H5NH2-HCl(s) => C6H5NH3+ (aq)+ Cl-(aq)
H2O is shown over the arrow.
What is the pH of the solution?
You have added a salt to water. The pH will be determined by the (H^+) of the solution from the hydrolyzed salt; i.e., reaction with water.
C6H5NH3^+ + HOH ==> H3O^+ + C6H5NH2
At the beginning (C6H5NH3^+) = 10g/molar mass/0.250 L.
(H3O^+) initially is zero (from the hydrolysis).
C6H5NH2 initially is zero.
What happens in the hydrolysis.
x molar C6H5NH3^+ hydolyzes; therefore, x mol C6H5NH2 is formed and x mol H3O^+ is formed. How much C6H5NH3^+ remains? It will be beginning molarity minus x.
Now write the equilibrium equation, plug in the values and solve for x.
Convert that to pH.
Ka = (C6H5NH2)(H3O^+)/(C6H5NH3^+)
Post your work if you get stuck.
I got that far but I do not have the Ka value, so I am stuck.
No problem. You can convert any Ka or any Kb with Kw = KaKb. Kw, of course, is 1E-14. You look up the Kb for aniline. There should be a table of ionization constants in your text or notes.
So I just plug the numbers into that equation, and then use the Ka to in the equilibrium constant equation, and then plug X in the equation for pH? Will that be the pH or the pOH?
Look at the equation. Did we let x stand for OH or H.
Thank you so much!!