To make a solution with a pH of 4.50 a student used the followingprocedure: A ce

Question

To make a solution with a pH of 4.50 a student used the followingprocedure: A certain amount of sodium acetate along with 0.300 moles ofacetic acid is added to enough water to make a solution of1.00L
How many grams of sodium acetate were added? Ka = 1.80 x 10-5 A certain amount of sodium acetate along with 0.300 moles ofacetic acid is added to enough water to make a solution of1.00L
How many grams of sodium acetate were added? Ka = 1.80 x 10-5

Explanation / Answer

The equation for the acetic acid buffer is: CH3COOH(aq) + OH-(aq) ---->CH3COO-(aq) + H2O(l) acetic acid        sodium acetic For this question, you can either setup an ICE table or use theHenderson-Hasselbach equation. I'm going to use the Hasselbach equation. If you rather use the ICEtable, look it up your the textbook to see how to do it. The Hasselbach equation is: pH = pKa + log([conjugate base]/[conjugate acid]) For your question: 4.50 = -log(1.80 x 10^-5) + log([moles/liters ofsodium acetic]/[0.300 moles/liters of acetic acid]) You want to find the unknown, which is sodium acetic. the -log(1.80 x 10^-5) is about the pH of 4.7, so: 4.50 = 4.7 + log([sodium acetic]/[0.300 moles/liters of aceticacid]) Subtract 4.7 from 4.50. -.2 = log([sodium acetic]/[0.300 M of acetic acid]).....(Rememberthat Molar Concentration (M) = moles/liter) To remove the log sign, multiple each side by the 10th power:10^(-.2) = [x]/[.300 M of acetic acid] solve for x by moving the denominator to the numerator: .18928 M =sodium acetate. After that find the moles of sodium acetate from solution of 1.00liter. use the moles to convert it into grams by multiplying sodiumacetate's molar mass (81.98g). I hope this helps. Check if there are any mathematical mistakes=).

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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