Asked by Anonymous
40 mL of 0.150 M ammonium hydroxide is added to 20 mL of 0.3 M perchloric acid. What is the final pH of the resulting solution?
Answers
Answered by
DrBob222
NH4OH + HClO4 ==> NH4ClO4 + H2O
millimols NH4OH = M x L = 6
mmols HClO4 = M x L = 6
So the acid and base exactly neutralize each other and you have the NH4ClO4 in aqueous solution at the end of the reaction. The concentration of the salt is mmols/mL = 6/60 = 0.1 M. The pH of the solution depends upon the hydrolysis of the NH4ClO4 salt. Only the NH4^+ hydrolyzes as
.......NH4^+ + H2O ==> NH3 + H3O^+
I......0.1..............0.....0
C......-x...............x.....x
E.....0.1-x.............x.....x
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.1-x) and solve for x = (H3O^+). Convert that to pH. Post your work if you get stuck. By the way, we don't use NH4OH anymore; i.e., it's NH3 + H2O so the Kb for NH3 is the same as the "old school" use of NH4OH.
millimols NH4OH = M x L = 6
mmols HClO4 = M x L = 6
So the acid and base exactly neutralize each other and you have the NH4ClO4 in aqueous solution at the end of the reaction. The concentration of the salt is mmols/mL = 6/60 = 0.1 M. The pH of the solution depends upon the hydrolysis of the NH4ClO4 salt. Only the NH4^+ hydrolyzes as
.......NH4^+ + H2O ==> NH3 + H3O^+
I......0.1..............0.....0
C......-x...............x.....x
E.....0.1-x.............x.....x
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.1-x) and solve for x = (H3O^+). Convert that to pH. Post your work if you get stuck. By the way, we don't use NH4OH anymore; i.e., it's NH3 + H2O so the Kb for NH3 is the same as the "old school" use of NH4OH.
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