Answer to Question #176410 in Inorganic Chemistry for SONA GEORGE

Question

) A buffer was prepared by mixing 1.00 mole of ammonia and 1.00 mole of ammonium chloride to form an aqueous solution with a total volume of 1.00 litre. To 500 mL of this solution was added 30.0 mL of 1.00 M NaOH. What will be the change in pH of this solution?

Answer 1

Number of mol of NH3 = 500/1000 = 0.5m

Number of moles of NH4Cl = 0.5 mol

Number of moles of NaOH = 30/1000 = 0.03

PKa of NH3 = 9.25

PH = PKa + log (NH3/NH4 +)

= 0.5/0.5 = 1

Log 1 = 0

PH = 9.25 - 0 = 9.25

∆ PH = 9.25 - 4.6 = 4.65