Question

# A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the...

A 75.0-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 13.0 mL of HNO3.

no of moles of NH3 in 75.0mL of 0.200M NH3 solution

= 0.075 x0.2 = 0.0150 mol NH3

no o fmoles of HNO3 in 13.0mL of 0.500M HNO3

= 0.013 x 0.500 = 0.0065 mol HNO3

net reaction is

NH4+ ---> NH3 + H+

1mol NH3 reacts with 1 mol HNO3 to produce 1 mol NH4NO3

moles of NH3 excess = 0.0150 - 0.0065 moles = 0.0085 moles

moles of NH4Br formed = 0.0065

total volume = 75 + 13 = 88 mL = 0.088 L

[NH3] = 0.0085 / 0.088 = 0.0966 M

[NH4+] = 0.0065 / 0.088 = 0.0738 M

from Kb calculate the pKb =

pKb = -logKb = -log(1.8×10−5) = 4.74

pOH = pKb + log([NH4] / [NH3]

pOH = 4.74 + log(0.0738 / 0.0966)

pOH = 4.74 + (-0.12)

pOH = 4.62

now use the formula

pH + pOH = 14

pH = 14-pOH

pH = 14-4.62

pH = 9.38

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