Question

You titrate 15 mL of a 0.10 M solution of methylamine (pkb = 3.36) with a titrant that is 0.10 M HCl. What is the pH at the midpoint (halfway point) of the titration? Select one or more: a. Cannot be determined from given information b.pH = 11.04 c.pH = 7.00 = d. pH = 10.64 e.pH = 3.36 You titrate a 25 ml sample of 0.10 M ammonia (Ko = 1.8 X 105) with 0.15 M HNO3. At the equivalence point the solution will be Select one or more: a. Not enough information to determine b. Acidic c. Basic d. Neutral

Answer 1

ANSWER 1

The midpoint is when the moles of strong acid added = ½ moles of base B

Actuly after calculation direct formula is-

pOH = pKb at mid point of titration

so, pOH = pKb = 3.36

=> pH = 14 - 3.36

= 10.64

ANSWER 2 .

FOrmula for equivalence point ==> pOH = 1/2 [ pKw + pKb + log C] ----------(EQN 1)

C is salt concentration ,  

now for calculation of C  

firstly we will calculate total volume and milimoles of base given

For NH3 N1 = 0.1 ,V1 =25 mli equivalent = N1V1 = 2.5

For HNO3 N2 = 0.15 , V2 =? mili equivalent = N2V2 => .15 V2

both mili equivalent will be same

so, N1V1 = N2V2

V2 = 2.5 / 0.15

V2 = 16.66 ML

total volume = V1+ V2 => 31.66

C = [SALT] = Mili moles of base / V1+ V2 Kb = 1.8 * 10^-5

= 2.5 / 31.66 pKb = -log Kb

  = .0789 pKb = 5- log 1.8

NOW  from EQN 1 pKb = 4.74

pOH = 1/2 [ 14 + 4.74 + log .0789]

= 1/2[18.74 - 1.1]

pOH = 8.82

=>pH = 14- 8.82

pH = 5.18 so, acidic in nature option b