Question

It is desired to neutralize a solution X that contains a mixture of potassium chloride and hydrobromic acid. Titration of 13.6 mL X with 0.107 M silver nitrate required 56.6 mL of the latter. The resulting precipitate, containing a mixture of AgCl and AgBr, was dried and found to weigh 1.090 g. How much 0.131 M sodium hydroxide should be used to neutralize 13.6 mL solution X?

Answer 1

Solution :-

AgCl molar mass= 143.32g/mol

AgBr molar mass = 187.77g/mol

Mass of precipitate = 1.090g

Volume of AgNO3 solution used for the titration = 56.6 ml = 0.0566 L

Molarity of the AgNO3 = 0.107 M

Lets first calculate the moles of AgNO3 used for the titration.

Moles = molarity x volume in liter

Moles of AgNO3 = 0.107 mol per L * 0.0566 L

                                = 0.006056 mol AgNO3

Br^+ Ag^+ --- > AgBr

Cl^- + Ag^+ --- > AgCl

Both have 1 :1 mole ratio with the Ag^+

Therefore total moles of AgNO3 used are same as the total moles of sum of the moles of Cl^- and Br^-

So lets assume moles of AgCl = x

Then moles of AgBr = (0.006056 – x)

(moles of AgCl * molar mass AgCl) + (moles of AgBr *molar mass of AgBr) = 1.090 g

(x*143.32 g per mol)+ ((0.006056 mol -x )*187.77 g per mol ) = 1.090 g

143.32 x + 1.137-187.77x = 1.090 g

143.32 x – 187.77 x = 1.090 g -1.137 g

-44.45 x = -0.047 g

X= -0.047 g/-44.45 g

X= 0.001057 mol

Therefore moles of AgCl are 0.001057 moles

Now using this moles of AgCl we can find the moles of AgBr

Moles of AgBr = total moles – moles of AgCl

                           = 0.006056 mol – 0.001057 mol

                           = 0.00499 mol AgBr

Using the moles of AgBr we can find the moles of HBr acid

1 mol AgBr= 1 mol HBr

Therefore moles of HBr are same as moles of AgBr that is 0.00499 mol HBr

HBr + NaOH ---- > NaBr + H2O

Using the moles of HBr we can find the moles of NaOH

0.00499 mol HBr * 1 mol NaOH / 1 mol HBr = 0.00499 mol NaOH

Now using the molarity and moles of NaOH we can find the volume of NaOH needed for the titration of 13.6 ml solution of X

Volume = moles / molarity

Volume of NaOH = 0.00499 mol / 0.131 mol per L

                                = 0.0381 L

Lets convert liter to ml

0.0381 L * 1000 ml / 1 L = 38.1 ml

Therefore we need 38.1 ml of NaOH