A 0.1997-g sample of primary standard sodium carbonate (Na2CO3) was weighed out

Question

A 0.1997-g sample of primary standard sodium carbonate (Na2CO3) was weighed out into a 250-mL beaker and diluted with 50 mL of distilled water. An HCl solution of unknown concentration was used to titrate the Na2CO3 in the beaker. After passing the first equivalence point at about 22 mL where the phenolphthalein indicator became colorless, a few drops of bromocresol green indicator were added. When the indicator began to change from blue to a green color, the titration was stopped and the solution was boiled and the indictor returned to a blue color as carbon dioxide was expelled from the solution. After cooling the sample back to room temperature, the titration was continued until a total of 45.12 mL of HCl had been added, and a distinct color change from blue to green was observed. Calculate the concentration of the HCl titrant solution using the volume to reach the 2nd equivalence point.

Explanation / Answer

moles of Na2CO3 = given mass / molar mass = 0.1997 g / 106 g/mol = 0.00188 mol

Molarity of Na2CO3 = 0.00188 / 0.05 L = 0.0376 M

Now,

HCl solution of unknown concentration was used to titrate the Na2CO3 in the beaker and first equivalence point is 22 mL.

At equivalence point

moles of Na2CO3 = moles of HCl

M(Na2CO3) * V(Na2CO3) = M(HCl) * V(HCl)

0.00188 mol = M(HCl) * 0.022 L

M(HCl) = 0.0854 M

nOW,

First equivalence point was observed at 22 mL and second equivalence point was observed at 45.12 mL.

At second equivalence point another 23.12 mL of HCl was added

So,

M(HCl) * V(HCl) = M(HCl)*V(HCl)

0.0854*0.022/0.0451 = M(HCl)

M(HCl) = 0.042 M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.