A student must make a buffer solution with a pH of 5.0. Determine which of the a
ID: 555586 • Letter: A
Question
A student must make a buffer solution with a pH of 5.0. Determine which of the acids and conjugate bases listed below are the best options to make a buffer at the specified pH. Weak acids: propionic acid, K2=1.34x10-5, 3.00 M ammonium citrate, Ka=4.06x10-7, 2.00 M acetic acid, K=1.75x10-5, 5.00 M formic acid, Ka=1.77x10-4, 2.00 M Conjugate bases: sodium propionate, CH3CH2COONa sodium formate, HCOONa sodium acetate trihydrate, CH3COONa• 3H20 sodium citrate dihydrate, C6H5O7Naz• 2H20 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 Based on this information, what volume of conjugate base should the student weigh out to acid should the student measure to make make the buffer solution with a pH=5.0? the 0.100 M buffer solution?Explanation / Answer
Ans. #1. The buffering capacity of a weak acid – conjugate base buffer is maximum when pKa of the weak acid is closest to the specified pH.
pKa of propionic acid = -log Ka = -log (1.34 x 10-5) = 4.87 (closest to pH 5.0)
pKa of ammonium citrate = -log Ka = -log (4.06 x 10-7) = 6.39
pKa of acetic acid = -log Ka = -log (1.75 x 10-5) = 4.76
pKa of formic acid = -log Ka = -log (1.77 x 10-4) = 3.75
Since pKa of propionic acid is closest to desired pH, it is the best choice for preparing the desired buffer.
#2. Conjugate base of propionic acid = Sodium propionate
A buffer consists of a weak acid and its respective conjugate base.
#3. Given, final concertation of weak acid, [AH] = 0.100 M
Volume of buffer = 100.0 mL = 0.100 L
Now, using Henderson- Hasselbalch equation for weak acid
pH = pKa + log ([A-] / [AH]) - equation 1
where, [A-] = conjugate base
[AH] = weak acid
Putting the values in equation 1-
5.0 = 4.87 + log ([A-] / [AH])
Or, [A-] / [AH] = antilog (5.0 – 4.87) = antilog (0.13)
Or, [A-] / [AH] = 1.349
Or, [A-] = 1.349 x [AH] ; [Given, [AH] = 0.100 ]
Or, [A-] = 1.349 x 0.100
Hence, [A-] = 0.1349
# Hence, [A-] = 0.1349 M
Required moles A- = Molarity x Volume of solution in liters
= 0.1349 M x 0.100 L
= 0.01349 mol
Required mass of conjugate base, Na-propionate = Moles x Molar mass
= 0.01349 mol x (96.061268 g/ mol)
= 1.296 g
#4. Required moles of Propionic acid = Molarity x Volume of solution in liters
= 0.100 M x 0.100 L
= 0.010 mol
Required mass of Propionic acid = 0.010 mol x (74.07944 g/mol) = 0.741 g
# Propionic acid is liquid at room temperature with density of 0.990 g/ mL
Required volume of propionic acid = Required Mass / Density
= 0.741 g / (0.990 g/ mL) = 0.748 mL
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