mols NH4Cl = grams/molar mass = about 0.005 but you need to do that more accurately. Then 0.005/0.115 = about 0.05
........NH4^+ + H2O ==>H3O^+ + NH3
I.......0.05M...........0......0
C........-x.............x......x
E......0.05-x...........x......x
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.05-x). Substitute from above and solve for x = (H3O^+) then convert to pH.
what is the pH of a solution prepared by adding 0.259g of ammonium chloride to 115mL of water? Kb of NH3 is 1.8x10^-5
1 answer
1 answer