Titration of acid and base (calculation)

 Titration is the method employed in volumetric analysis. In this method, an unknown concentration of a liquid substance is titrated against another of known concentration. We are going to learn how to calculate the concentration of the substance using the question below.

A is a dilute tetraoxosulphate (VI) acid. B contains 2g of sodium hydroxide per 250 cm3 of solution. 25cm³ of B requires 23cm³ of A for complete neutralization. Calculate

i.                     Concentration of solution B in mol/dm³

ii.                   Concentration of A in mol/dm³

iii.                  The number of hydrogen ions in 1dm³ of solution A

(Na=23, H=1, S=32, O=16)

Equation of reaction

H₂SO_4(aq)  +  2NaOH_(aq) → Na₂SO_4(aq)  + 2H₂O_(l)

Solution

i.                     Concentration of solution B in mol/dm³

250cm³   = 2g

1000cm³ = 2/250 x1000 = 8g

Molar mass of NaOH = 40g/mol

Concentration in mol/dm³   = 8/40   = 0.2mol/dm³

Or

C = n/V

Where C = concentration in mol/dm³

             .n = amount in mole

              V = volume in dm³

First convert 250cm³ to dm³ 

250 /1000 = 0.25dm³

Amount of NaOH  = 2/40  =  0.05mole

C = 0.05/0.25 = 0.2mol/dm³   

ii.                   Concentration of A in mol/dm³ 

 

C_aV_a/C_bV_b  = a/b

 

C_a  = ??

V_a  = 23cm³

C_b  = 0.2mol/dm³

V_b= 25cm³

.a  =  1

.b = 2

 

C_a x 23/0.2 x25 = 1/2.

 

C_a = 0.2 x 25/46 = 0.109mol/dm³

 

iii.                  Number of hydrogen ions in 1 dm³  of solution A

 

H₂SO₄   → 2H⁺   + SO₄^2-

  1          :     2

0.109    = 2x 0.109 = 0.22mole