Determination of percentage of an unknown salt by titration A volume of 11.19 mL

Question

Determination of percentage of an unknown salt by titration A volume of 11.19 mL of 0.1000 M NaOH solution was used to titrate a 0.553 g sample of unknown containing HC3HsO3 What is the molecular mass of HC3H503? (report answers to 4 or 5 significant figures) Submit Answer Tries 0/99 What is the percent by mass of HC3H503 in the unknown? Submit Answer Tries 0/99 In this problem what mass of sample in grams would be needed to deliver about 22.40 mL in the next trial? Submit Answer Tries o/99 In the second trial above, exactly 1.098 g was transferred into a flask to be titrated. If the initial buret reading is 0.10 mL, predict what the final buret reading be. Submit Answer Tries 0/99

Explanation / Answer

NaOH (aq) + HC3H5O3(aq) ? C3H5O3Na (aq)

From equation we can see, 1 mol of NaOH neutrlises 1 mol of HC3H5O3.

a) molecular mass = (3x 12 + 1 x 6 + 3x 16) g mol-1

= 90 g mol-1

b)

(0.01119L) x (0.1 mol NaOH/L) x (1 mol HC3H5O3 / 1 mol NaOH) x (90 g HC3H5O3/mol) = 0.1007 g HC3H5O3

So, percent by mass = (0.1007 g) / 0.553) x 100 = 18.2%

c) since, for 11.9 mL sample needed = 0.553 g

For 1 mL sample needed = 0.553 / 11.19 = 0.049 g

So, for 22.4 mL sample needed = 0.046 x 22.4 = 1.098 g

d)

since for second trial it requires 22.4 mL of the titrant, so final burette reading = (22.4 + 0.1) mL = 22.5 mL

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

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