Question

20 ml of a solution containing equal moles of n a 2 c o 3 in any nahco3 required 16 ml of 0.16m hcl solution to the phenolphthalein end point . what volume of a 0.10m h2so4 solution would have been required and methyl orange been used as indicator

Answer 1

Solution.

$$\frac{x(Na2CO3)}{x(NaHCO3)} = \frac{N(HCl)*V(phen)*E(Na2CO3)*2*100*1000*m}{1000*m*N(H2SO4)*(V(met)-2*V(phen)*E(NaHCO3)*100}$$

$$\frac{x(Na2CO3)}{x(NaHCO3)} = 1:1$$

$$1 = \frac{N(HCl)*V(phen)*E(Na2CO3)*2}{N(H2SO4)*(V(met)-2*V(phen))*E(NaHCO3)}$$

V(met) = 64.3 ml

Answer:

 V(met) = 64.3 ml