10.0 ml of Na 2 CO 3 solution is titrated against 0.2 M HCl solution. The following titre values were obtained in 5 readings. 4.8 ml , 4.9 ml , 5.0 ml , 5.0 l and 5.0 ml Based on these readings, and convention of titrimetric estimation of concentration of Na 2 CO 3 solution is. mM (Round off to the Nearest integer)

Question

10.0 ml of Na 2 CO 3 solution is titrated against 0.2 M HCl solution. The following titre values were obtained in 5 readings. 4.8 ml , 4.9 ml , 5.0 ml , 5.0 l and 5.0 ml Based on these readings, and convention of titrimetric estimation of concentration of Na 2 CO 3 solution is. mM (Round off to the Nearest integer) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Most precise volume of HCl =5 ml at equivalence point Meq. of Na 2 CO 3= meq. of HCl. Let molarity of Na 2 CO 3 solution = M, then M × 10 × 2=0.2 × 5 × 1 M =0.05 mol / L =0.05 × 1000 =50 mM

Most precise volume of HCl=5ml at equivalence point Meq. of Na2​CO3​= meq. of HCl. Let molarity of Na2CO3​ solution =M, then M×10×2=0.2×5×1 M=0.05mol/L =0.05×1000 =50mM

Most precise volume of $H Cl = 5 m l$

at equivalence point

Meq. of $N a_{2} C O_{3} =$ meq. of $H Cl$ .

Let molarity of $N a 2 C O_{3}$

solution $= M$ , then

$M \times 10 \times 2 = 0.2 \times 5 \times 1$

$M = 0.05 m o l / L$

$= 0.05 \times 1000$

$= 50 m M$