Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8×10−5) with 0.20 M HNO3. Calculate the pH...

Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8×10−5) with 0.20 M HNO3. Calculate the pH after addition of 50.0 mL of the titrant.

Expert Solution

NH3 millimoles = 50 x 0.20 = 10

millimoles of HNO3 = 50 x 0.20 = 10

NH3 + HNO3 -------------------> NH4+NO3-

10 10 0 -------------------> initial

0 0 10 -------------------> after reaction

here salt NH4+NO3 only remains in the mixture.

so pH is decided by salt hydrolysis of weak base strong acid salt

now

pKb of NH3 = 4.75

molarity of salt = millimoles / total volume

= 10 / (50+50)

= 0.1 M

C= 0.1 M

pH = 7 - 1/2 [pKb + logC]

pH = 7 - 1/2 [4.74 + log 0.1]

pH = 5.13

queen_honey_blossom answered 3 months ago

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