Question

calculate the ph at 25°C of 199mL of a buffer solution that is 0.440M NH4Cl and 0.44M NH3 before and after the addition of 1.5mL of 6.0M HNO3. The pKa for NH4+ is 9.75.
please I need the answer asap.....
pH after addition is not 9.16

Answer 1

Before addition of HNO3

pKa = 9.75

pKb = 4.25

pOH = pKb + log [NH4Cl / NH3]

pOH = 4.25 + log (0.44 /0.44)

pOH = 4.25

pH + pOH = 14

pH = 9.75 ---------------------------> answer

after addition of C millimoles of HNO3

C = 1.5 x 6.0 = 9

millimoles of NH4Cl = 0.440 x 199 = 87.56

millimoles of NH3 = 0.440 x 199 = 87.56

new pH :

on addition of C moles acid to basic buffer base moles decreases and acid moles increases

so

pOH = pKb + log [NH4Cl + C / NH3 -C]

= 4.25 + log [87.56 + 9 / 87.56 - 9]

= 4.34

pH + pOH =14

pH = 9.66------------------------> answer