Question

Concentrated reagent grade ammonia is reported as 14.8 F. However,it can lose strength rapidly if it is not tightly stoppered, and itshould be standardized if needed for precise work. HCl isstandardized against primary standard sodium carbonate: 0.2500 grams of sodium carbonate required 40.00 mL of HCl to reach theendpoint. 10.00 mL of concentrated ammonia is diluted to 1.000 liter and 25.00 mL of the diluted ammonia is titrated with the HCl: If the titration of the ammonia required 26.39 mL of HCL, what is the formality of the concentrated ammonia

Answer 1

m = 0.25 g of Na2CO3

V = 40 mL of HCl

2 mol of HCl = 1 mol o Na2CO3

mol of Na2CO3 = mass/MW = 0.25/105.9884 = 0.002358 mol

HCl mol = 2*Na2CO3 = 2*0.002358 = 0.004716 mol of HCl

[HCl] =mol/V = 0.004716 /0.04 = 0.1179 M

now...

Volume of HCl used:

26.39 mL

mol o fHCl = MV = 0.1179 * 26.39/1000 = 0.003111 mol of HCl used

mol of NH3 = ml of HCl due to 1:1 ratio

0.003111 mol of HCl used = 0.003111 mol of NH3

Now...

0.003111 mol of NH3 present in V = 25 mL of dilution

[NH3] = mol/V = 0.003111 / (25*10^-3) = 0.12444 mol of NH3

note that

molarity of 25 mL = molarity o 1 L = 1000 mL

then...

M1*V1 = M2*V2

0.12444*1000 = M2*10

M2 = 0.12444*100

M2 = 12.444 M for NH3