Calculate the pH after addition of 16.36 mL of 0.1349 M HBr to 25.00 mL of 0.1444 M NH3. Kb(NH3) = 1.800e-5.
Answer: 9.059
PLEASE SHOW STEPS!
Use the H-H equation.
Technically one should use concentrations but I like to work in millimols = mL x M. And since both numerator and denominator for M = millimols/mL and mL is the same (because it's the same solution) then using millimoles gives the same answer.
mmols HBr = 16.36 x 0.1349 = approx 2.2
mmols NH3 = 25.00 x 0.1444 = approx 3.6
The rxn produces a buffered solution.
........HBr + NH3 ==> NH4Br
I........0.....3.6......0
add.....2.2...............
C......-2.2...-2.2.....2.2
E........0.....1.4.....2.2
Note that the solution is that of a weak base and its salt, thus a buffered solution.
pH = pKa + log (base)/(acid)
pH = pKa + log (1.4/2.2)
Plug and chug. Note: You should confirm those calculations since they are just close estimates. Use more accurate numbers when you do it. To find pKa, you have Kb so pKb = -logKb, then
Kb*Ka = Kw = 14. You know Kw and Kb, solve for Ka and use that in the HH equation.