Asked by Ana
Ammonia is a weak base with a Kb of 1.8 x 10-5. A 100.0 mL sample of 0.13 M aqueous solution of ammonia (NH3) is mixed with 87 mL of 0.15 M solution of the strong acid HCl, calculate the pH of the final solution.
A) 2.9
B) 11.1
C) 8.8
D) 5.2
E) None of the above
A) 2.9
B) 11.1
C) 8.8
D) 5.2
E) None of the above
Answers
Answered by
DrBob222
millimols NH3 = mL x M = 100 x 0.1 = 13
mmols HCl = 87 x 0.15 = 13.05
My best guess is that these are to be considered the same since the 87 has two s.f. and 13 does too.
........NH3 + HCl ---> NH4Cl
I.......13....13........0
C......-13...-13.......+13
E........0.....0.........0
I'm assuming the HCl exactly neutralizes the NH3 in which case the pH is determined by the hydrolysis of the salt, NH4Cl, which has a concn of 13mmols/187 mL = about 0.07 and you need to clear up that number a little as well as all that folllow.
........NH4^+ + H2O ==> NH3 + H3O^+
I.......0.07............0......0
C.......-x..............x......x
E......0.07-x...........x......x
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.07-x) and solve for x = (H3O^+ and convert to pH. NH4^+ salt have a pH of about 5.2
mmols HCl = 87 x 0.15 = 13.05
My best guess is that these are to be considered the same since the 87 has two s.f. and 13 does too.
........NH3 + HCl ---> NH4Cl
I.......13....13........0
C......-13...-13.......+13
E........0.....0.........0
I'm assuming the HCl exactly neutralizes the NH3 in which case the pH is determined by the hydrolysis of the salt, NH4Cl, which has a concn of 13mmols/187 mL = about 0.07 and you need to clear up that number a little as well as all that folllow.
........NH4^+ + H2O ==> NH3 + H3O^+
I.......0.07............0......0
C.......-x..............x......x
E......0.07-x...........x......x
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.07-x) and solve for x = (H3O^+ and convert to pH. NH4^+ salt have a pH of about 5.2