Calculate the molarity of a solution of NH3 in water with a PH of 11.17
2 answers
I don't know
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.