Sample Problem. standard Na2CO3 required tion with A 0.1355 g sample (uncorrecte

Question

Sample Problem. standard Na2CO3 required tion with A 0.1355 g sample (uncorrected mass) of dry, primary 25.05 mL of HCi solution (at 23.3 °C) to reach the end point. A the same amount of indicator required 0.04 mL. Use linear interpolation an data in Table 2-7 of Harris to estimate the density of water at 23.3 °C. What concentration of the HCI solution, corrected to 20 °C? blank titra is the [Buoyancy correction factor: mrue/mread 1.000 3244] [Interpolated density of water at 23.3 °C: 0.997 468 9 g/mL] [HCI concentration at 20 °C: 0.102343 M] im Mm10003 mread lon a4 -23 0.975A1

Explanation / Answer

The sample contains

0.1355 g / 105.988 g/mol = 1.2784x10-3 mol Na2CO3

and will be titrated with 2x 1.2784 mol = 2.5569x10-3 mol HCl.

Correct the HCl volume to 20 °C

   25.05 mL x 0.99747g/mL / 0.9982 g/mL = 25.03 mL

The actual concentration of HCl is                

2.5569x10-3 mol / 25.03x10-3 L = 0.10215 mol/L

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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