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
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
We have 500.0 mL of a buffer solution that is 0.636 M in NH3 and 0.345 M in NH4Cl. We add 500.0 mL of a solution 0.249 M in HCl. The final volume is 1.0000 L. The Kb value for NH3 is 1.8 x 10-5. What is the pH of the final solution?
Calculate the change in pH when 3.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq). NH3 Kb=1.8x10^-5 Calculate the change in pH when 3.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.
Calculate the pH of a solution prepared by adding 20.0 mL of 0.100 M HCl to 80.0 mL of a buffer that is comprised of 0.25 M NH3 and 0.25 M NH4Cl. Kb of NH3 = 1.8 x 10-5. ОА. 9.17 ОВ. 4.83 0 o С. 9.34 OD.9.26 OE. 4.66
What amount of NH4Cl must be added to 50.0 mL of 0.30 M NH3 in order to produce a buffer of pH 9.00? Assume the addition of NH4Cl does not change the volume of the solution. Kb = 1.8 × 10-5 for NH3
What is the pH of the solution from adding 25 mL of 0.33 M HCl to 25 mL of 0.58 M NH3? (Kb for NH3 is 1.8 x 10^-5)
How many mL of a 4.00 M HCl solution must be added to 250 mL of a 0.250 M NH3 solution to make a buffer with pH = 9.10? Look up Ka or Kb in a suitable source.
What is the pH of a buffer formed from 50 mL of 15.0 M NH3 and 53.5 g of NH4Cl in enough water to make 500 mL of solution? (Kb = 1.8 x 10-5)
How many grams of dry NH4Cl need to be added to 2.50 L of a 0.500 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.74? Kb for ammonia is 1.8*10^-5.
How many grams of dry NH4Cl need to be added to 1.70 L of a 0.100 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.62? Kb for ammonia is 1.8×10−5.
How many grams of dry NH4Cl need to be added to 1.50 L of a 0.500 M solution of ammonia, NH3,to prepare a buffer solution that has a pH of 8.79? Kb for ammonia is 1.8*10^-5.