Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

3/21/2018 08:00 AM A 13.8/15 Gr Calculator -Periodic Table Question 13 of 13 Map

ID: 1028177 • Letter: 3

Question

3/21/2018 08:00 AM A 13.8/15 Gr Calculator -Periodic Table Question 13 of 13 Map A student must make a buffer solution with a pH of 5.5. Determine which of the acids and conjugate base listed below are the best options to make a buffer at the specified pH. Weak acids: O ammonium citrate, K 4.06x107 O acetic acid, K-1.75x10-5 formic acid, Ka= 1.77x10-4 O propionic acid, Ka-1.34x105 Conjugate bases: O sodium acetate trihydrate CHacOONa- 3H2o O sodium propionate CHaCH-COONa O sodum citrate dihydrate CoHsONa 2HO O sodium formate HCOONa Scroll down to view more The final volume of buffer solution must be 100.00 mL and the final concentration of the weak acid must be 0.100 M. Based on this information what mass of solid conjugate base should the student weigh out to make the buffer solution with a pH 5.5? Number grams Hint Previous check Answer 0 Next Exit

Explanation / Answer

The first thing to do is to calculate the Pka of the given acids

PKa = - log Ka

using the information provided

ammonium citrate Pka = 6.39

acetic acid Pka = 4.757

formic acid Pka = 3.75

propionic acid Pka = 4.87

the most suitable choice should be 4.87 since it is closer to 5.5

the conjugate base will be sodium propionate

Part 2

Hendersson hasselbach equation

PH = Pka + log (propionate ion / propionic acid)

5.5 = 4.87 + log ( A- / 0.1)

5.5 = 4.87 + log (A / 0.1)

0.63 = log (A / 0.1)

4.2658 = A / 0.1

A = 0.42658 M

The total volume is 100 ml so

Moles = Molarity * volume

moles = 0.42658 * 0.1 = 0.042658 moles

molar mass of sodium propionate 96.07 g/mol

mass = moles * molar mass = 0.042658 * 96.07 = 4.09 grams of sodium propionate

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote