Question

A) What volume of 0.456 M HNO3 is needed to titrate 75.00 mL of 1.0x10-3 M Ca(OH)2 to the equivalence point? B) What is the pH of a buffer system made by dissolving 10.70 grams of NH4Cl and 2.0 mL of 12.0 M NH3 in enough water to make 1.00 L of solution? (Kb=1.8x10-5 for NH3)

Answer 1

A) 2HNO3 + Ca(OH)2 -------> Ca(NO3)2 + H2O

Molarity of Ca(OH)2 solution = 0.001M

Volume of Ca(OH)2 solution = 75ml

No of mole of Ca(OH)2 = (0.001 mol/1000ml )× 75ml =7.5×10^-5mol

7.5× 10^-5mol of Ca(OH)2 require 1.5× 10^-4mol of 2HNO3

molarity of HNO3 = 0.456M

Volume HNO3 solution having 0.00015mol of HNO3 = (1000ml/0.456mol)×0.00015mol = 0.33ml

So, the answer is 0.33ml

2) Mass of NH4Cl = 10.70g

molar mass of NH4Cl = 53.49g/mol

No of mole NH4Cl = 10.70/53.49g = 0.2000

Molarity of NH3 solution = 12M

Volume of NH3 solution = 2ml

No of mole of NH3 = (12mol/1000ml)×2=0.024mol

Total volume buffer solution = 1Litre

Concentration of NH4+ = 0.2M

Concentration of NH3 = 0.024M

Henderson Hasselbalch equation

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

Kb of NH3 = 1.8×10^-5

pKb = - log ( Kb)

= - log ( 1.8×10^-5)

= 4.74

pOH= 4.74 + log (0.2/0.024)

= 4.74 + 0.92

= 5.66

pH + pOH = 14

Therefore,

pH = 14 - 5.66

= 8.34

So, the answer is 8.34