Asked by Nsemeke Edet
Calculate the molarity of a solution of NH3 in water with a PH of 11.17
Answers
Answered by
Nsemeke Edet
I don't know
Answered by
DrBob222
pH = 11.17 = - log (H^+)
(H^+) = 6.76E-12
(H^+)(OH^-) = 1E-14
(OH^-) = 1E-14/6.76E-12 = 0.00148 M
.....................NH3 + H2O ==> NH4^+ + OH^-
I......................Y......................0..............0
C....................-x......................x..............x
E..................Y-x......................x..............x
Kb for NH3 = (NH4^+)(OH^-)/(NH3) = (x)(x)/(Y-x)
Look up Kb in your tables. It will be about 1.8E-5 but use whatever is in your text/notes/or what you find on the Internet. You know x = 0.00148. Solve for Y. That will be the molarity of the NH3 in the solution. Post your work if you get stuck.
(H^+) = 6.76E-12
(H^+)(OH^-) = 1E-14
(OH^-) = 1E-14/6.76E-12 = 0.00148 M
.....................NH3 + H2O ==> NH4^+ + OH^-
I......................Y......................0..............0
C....................-x......................x..............x
E..................Y-x......................x..............x
Kb for NH3 = (NH4^+)(OH^-)/(NH3) = (x)(x)/(Y-x)
Look up Kb in your tables. It will be about 1.8E-5 but use whatever is in your text/notes/or what you find on the Internet. You know x = 0.00148. Solve for Y. That will be the molarity of the NH3 in the solution. Post your work if you get stuck.
There are no AI answers yet. The ability to request AI answers is coming soon!