should contain more than4% ac de acid (CHCOOH, molar mass 60.0 g mol of this sol

Question

should contain more than4% ac de acid (CHCOOH, molar mass 60.0 g mol of this solution are titrated 25.0 ml of vinegar are diluted to 100.0 ml and 10.0 mL samples against 0.100 mol L- NaOH solution. An average of 18.1 mL NaOH are required to ne the vinegar. You will need to calculate the concentration of the diluted solution and hence the percentage concentration of acetic acid in the original vinegar via the following questions: CH,COOH(aq) + NaOH(a) CHCOONa(aq) + H2O(l) Is the known substance NaOH or acetic acid? Find the number of moles of the known substance a. According to the balanced reaction, what is the mol-to-mol ratio of acetic acid to sodium hydroxide? b. Use your answer in part B to determine how many moles of the unknown substance were utilized in this titration. c.

Explanation / Answer

a)

the "known" substance is the NaOH, which is standardized and we know its volume + concentration

the "unkown" must be the acetic acid present in the vinegar

mol of NaOH = known = MV = 0.1*18.1*10^-3 = 0.00181 mol of NaOH used

b)

ratio is 1:1 so

1 mol of acid = 1 mol of base

0.00181 mol of base = 0.00181 mol of acid

c)

moles of acid, from 1:1 ratio was = 0.00181

d)

find concentration in diluted sample

mol of acid = 0.00181

Vacid = 10 mL

[Acid] diluted = mol/V = 0.00181/(10*10^-3) = 0.181 M

this is also the same concentration in the V = 100 mL

e)

get undiluted value

C1V1 = C2*V2

C1 * 25 mL = 100 mL * 0.181 M

C1 = 100/25 * 0.181

C1 = 0.724 M

the formula is "dilution law"

f)

convert to g/L

0.724 mol per liter * 60 g /mol = 43.44 g / L

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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