Question

if 155 mL of 0.043 M HCl solution is added to 235 mL of a buffer solution which is 0.33 M in NH3 and 0.56 M in NH4Cl, what will be the pH of the new solution?

the Kb of NH3 is 1.8*10^-5

Answer 1

Sol .

As initial millimoles of NH3 = Conc. of NH3 × Volume of buffer

= 0.33 × 235 = 77.55 mmol

and , initial millimoles of NH4Cl = Conc. of NH4Cl × Volume of buffer = 0.56 × 235 = 131.6 mmol

Now , millimoles of HCl added = Conc. of HCl × Volume of HCl = 0.043 × 155 = 6.665 mmol

So , after addition of HCl , millimoles of NH3 decreases and millimoles of NH4Cl increases .

So ,

New millimoles of NH3 = 77.55 - 6.665 = 70.885 mmol

New millimolea of NH4Cl = 131.6 + 6.665 = 138.265 mmol

Now , Kb = 1.8 × 10^-5  

So , pKb = - log (1.8 × 10^-5 ) = 4.74  

So , Using Henderson - Hasselbalch equation ,

pOH = pKb + log ( New millimoles of NH4Cl / New Millimoles of NH3 )

pOH = 4.74 + log ( 138.265 / 70.885 )

= 5.03

Therefore , pH = 14 - pOH = 14 - 5.03 =  8.97