Question

A 0.1355 g sample of dry, primary-standard Na2CO3 required 25.05 mL of HCl solution (at 23.3°C)...

A 0.1355 g sample of dry, primary-standard Na2CO3 required 25.05 mL of HCl solution (at 23.3°C) to reach the end point. A blank titration with the same amount of indicatio required 0.04 mL. What is the concentration of the HCl solution at room temperature.

I think the answer should be .1022673 M but I'm having trouble figuring out the process of getting there. Please help.

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Homework Answers

Answer #1

Na2CO3 + 2HCl ------> 2NaCl + CO2 +H2O

1 mole     2 moles

no of moles of Na2CO3 = W/G.M.Wt

                                  = 0.1355/106 =0.001278 moles

from the balanced equation

1 mole of Na2CO3 react with 2 moles of HCl

0.001278 moles of Na2CO3 react with = 2*0.001278/1 =0.002556 moles of HCl

molarity = no of moles/volume of solution L

volume of solution = 25.05ml = 0.02505 L

              = 0.002556/0.02505 = 0.102035 M

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