A75.0-ml volume of 0.200 m nh3 (kb=1.8×10−5) is titrated with 0.500 m hno3. calculate the ph after the addition of 28.0 ml of hno3.

A) Initial moles of NH3 = 75/1000 x 0.200 = 0.015 mol

Moles of HNO3 added = 28/1000 x 0.500 = 0.014 mol

NH3 + HNO3 => NH4+ + NO3-

Moles of NH3 left = 0.015 - 0.014 = 0.001 mol

Moles of NH4+ = 0.014 mol

Ka(NH4+) = Kw/Kb(NH3)

= 10-14/1.8 x 10-5 = 5.556 x 10-10

Henderson-Hasselbalch equation:

pH = pKa + log([NH3]/[NH4+])

= -log Ka + log(moles of NH3/moles of NH4+) since volume is the same for both

= -log(5.556 x 10-10) + log(0.0065/0.014)

= 8.14

Answer : The pH after the addition of 28.0 ml of is, 8.1

Explanation :

First we have to calculate the moles of and .

The balanced chemical reaction is,

Moles of left = Initial moles of - Moles of added

Moles of left = 0.015 - 0.014 = 0.001 mole

Moles of = 0.014 mole

Now we have to calculate the .

Now we have to calculate the

Now we have to calculate the pH by using Henderson-Hasselbalch equation.

Now put all the given values in this expression, we get:

Therefore, the pH after the addition of 28.0 ml of is, 8.1

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