Question

used a 0.25 M HCl solution.

Dried primary standard grade Na₂CO₃will be available.  Accuratelyweigh out 0.2564 g portion of Na₂CO₃into a 250 mL Erlenmeyer flask with 50 ml of water. Total amount of HCL used in titration was 17.7 ml what is the new concnetration of HCL in the solution.

Answer 1

Volume of Na₂CO₃ solution = 50.0 mL

Volume of HCl added = 17.7 mL

Total volume = 67.7 mL

According to dilution law,

M1 * V1 = M2 * V2

where M1 = molarity before dilution

V1 = volume before dilution

M2 = molarity after dilution

V2 = volume after dilution

For HCl , M1 = 0.25 M , V1 = 17.7 mL , V2 = total volume = 67.7 mL

M2 = (M1 * V1) / V2

M2 = (0.25 M * 17.7 mL) / 67.7 mL

M2 = 0.25 M * (17.7 mL / 67.7 mL)

M2 = 0.25 M * 0.261

M2 = 0.06536 M

New concentration of HCl = 0.065 M