Question

HW 11 - Fall 19 YANG > Activities and Due Dates > HW 11 Resources Feedback ignment Score: 150/1800 Question 1 of 18 > A student must make a buffer solution with a pH of 5.00 Determine which weak acid is the best option to make a buffer at the specified pH. O propionic acid, K, -1.34 x 10- 3.00 M O acetic acid, K, -1.75 x 10-5, 5.00 M O formic acid, K, = 1.77 x 10-4 2.00 M O ammonium citrate, K. - 4.06 x 10-7.2.00 M Determine which conjugate base is the best option to make a buffer at the specified pH. sodium citrate dihydrate, CHO Na 2HO sodium propionate. CH CH.COONa R F G H J

Answer 1

1.

Part a

Using Henderson -Hasselbalch equation

pH of buffer expressed as

pH = pKa + log [A^-]/[HA]

Buffer is most effective when [A^-]/[HA] = 2.

Hence, pH = pKa.

Hence, to prepare buffer at pH = 5

The pKa of the acid should be close to 5.

Given, Ka of propionic acid = 1.34×10^-5

Hence, pKa = 4.87.

Ka of acetic acid = 1.75×10^-5

Hence, pKa = 4.75

Ka of formic acid = 1.77×10^-4

pKa = 3.75

Ka of ammonium citrate = 4.06×10^-7

pKa = 6.33.

Therefore, propionic acid is best option.

Part b

Buffer is made of weak acid and its conjugate base.

Therefore sodium propionate is used prepare the buffer.

Part c.

mmoles of acid in 100 mL solution

= 100 × 0.100 = 10

Let mmoles of conjugate base = x

pH = pKa + log [conjugate base]/[acid]

5.00 = 4.87 + log ( x/10)

Or, log(x/10) = 0.13

Or, (x/10) = 10^0.13

Or, (x/10) = 1.35

Or, x = 13.5

Now, moles of conjugate base needed = (13.5/1000)

= 0.0135

So, mass of conjugate base needed

= moles of conjugate base ( sodium propionate)× molar mass of sodium propionate

= 0.0135 (mol) × 96 (g/mol)

= 1.296 g.

Part d.

Initial molarity of the acid = 3.00 M

Final concentration = 0.1 M

So, moles of acid

= M₁V₁ = M₂V₂ ​​​​​​

Or, 3.00 × V₁ = 0.1 × 100

Or, V₁ = (10/3) = 3.33 mL.

So, take 3.33 mL acetic acid and dilute it to 100 mL.