Answer to Question #253991 in Physical Chemistry for Chemistry

Question

A solution made from pure Potassium hydroxide contained 1.85 g of Potassium hydroxide in exactly 200 cm3 of water. Using phenolphthalein indicator, titration of 20.0 cm3 of this solution is carried out v/s sulphuric acid. 9.35 cm3 of sulphuric acid solution is required for complete neutralisation. [atomic masses: K= 39, O = 16, H = 1] (a) write the equation for the titration reaction. (b) calculate the molarity of the Potassium hydroxide solution. (c) calculate the moles of Potassium hydroxide neutralised. (d) calculate the moles of sulphuric acid neutralised. (e) calculate the molarity of the sulphuric acid

Answer 1

(a) H₂SO₄ + 2KOH----> K₂SO₄ + 2H₂O.

(b)Moles of KOH in 1.85g=$\frac{mass}{Rfm}$

     =$\frac{1.85}{56}$

     =0.033moles.

But moles=molarity × volume÷1000

     0.033=molarity×200/1000

     Hence,molarity=5 ×0.033

         =0.165M.

(c)moles neutralized=molarity ×volume/1000

   = 0.165 X $\frac{20}{1000}$

   =0.0033moles.

(d)moles of acid is $\frac{1}{2}$ moles of base,since mole ratio from eqn is 1:2.

Hence moles of acid=$\frac{1}{2}$ ×0.0033

    =0.00165moles

(e)moles=molarity× volume/1000

  0.00165=mol×$\frac{9.35}{1000}$

Hence,mol=(0.00165× 1000)÷9.35

   =0.1765 M.