What concentrations of acetic acid (pKa 4.76) and acetate would be required to p

Question

What concentrations of acetic acid (pKa 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 4.9? Note that the concentration and/or pH value may differ from that in the first question. STRATEGY 1. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid), LA VIHA] 2. Use the mole fraction of acetate to calculate the concentration of acetate 3. Calculate the concentration of acetic acid. That is, there are 1.4 molecules acetate Step 1: The ratio of base to acid is T 1.4 of acetate for each molecule of acetic acid acetic acid Step 2: Use the mole fraction of acetate to calculate the concentration of acetate Number acetate M

Explanation / Answer

1.

Using Henderson Hasselbach

pH= pKa+ log ([salt]/[acid])

4.9=4.76+log ([salt]/[acid])

log ([salt]/[acid]) = 4.9 - 4.76=0.14

So ([salt]/[acid]) = 100.14 = 1.38

ratio of concentration acetate /acid =1.38

Total concentration of acetate +acid =0.15

If concentrations of acetic acid is x , concentration of acetate is1.38x.

So,

x + 1.38x = 0.15

2.38x =0.15

x = 0.063

acetic acid = x = 0.063 M and

acetate = 0.15 - 0.063 = 0.087 M

2.

Using Henderson Hasselbach

pH= pKa+ log ([salt]/[acid])

4.5=4.76+log ([salt]/[acid])

log ([salt]/[acid]) = 4.5 - 4.76 = - 0.26

So ([salt]/[acid]) = 10-0.26 = 0.55

ratio of concentration acetate /acid = 0.55

Total concentration of acetate +acid =0.10

If concentrations of acetic acid is x , concentration of acetate is 0.55x.

So,

x + 0.55x = 0.10

1.55x =0.10

x = 0.065

acetic acid = x = 0.065 M and

acetate = 0.10 - 0.065 = 0.035 M

3.

Molarity = Moles / Liter

Volume of NaOH = 16.5 mL = 0.0165 L

Molarity = 0.280 M

So, moles of NaOH = 0.280 M x 0.0165 L = 0.00462 moles

Since the acid is diprotic acid, so to reach second equivalence point,

2 moles of NaOH will react with 1 mole of acid.

1 moles of NaOH will react with 1/2 mole of acid.

0.00462 moles of NaOH will react with 0.00462/2 mole of acid.

0.00462 moles of NaOH will react with 0.00231 mole of acid.

So,

0.00231 mole of acid = 0.354 g

1 mole of acid = (0.354 / 0.00231) g = 153.25 g

So, the molar mass of the acid = 153.25 g/mol = 1.53 x 102 g/mol

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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