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

A student must make a buffer Solution with a pH of 5.0. Determine which of the a

ID: 502802 • 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 Incorrect. O ammonium citrate, Ka-4.06x10 2.00 M Scroll down for more O acetic acid, Ka 1.75x10-5, 5.00 M o propionic acid, Ka 1.34x10 5, 3.00 M O formic acid, Ka-1.77x10 2.00 M Conjugate bases: O sodium formate, HCOONa sodium propionate, CH3CH2COONa O sodium citrate dihydrate, C6H507Na3. 2H20 sodium acetate trihydrate O CH3 COONa. 3H20 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 th n

Explanation / Answer

given that;

pH = 5.0

for making best buffer of pH = 5.0 we need the weak acid that has nearst pKa

thus we used, propionic acid pKa (4.87 ) is nearer to given pH (5.0) value. so it is suitable

for making best buffer at pH = 5.0

propionic acid, Ka=1.34x10^–5

pKa of propionic acid = 4.87

the corresponding salt formed from the propanoinc acid is sodium propionate CH3CH2COONa

pH = pKa + log [salt / acid]

5.0 = 4.87 + log [salt / acid]

[salt / acid] = 1.349

acid moles = 100 x 0.1 / 1000

                  = 0.01

salt / acid = 1.349

salt / 0.01 = 1.349

salt (conjugate base) moles = 1.349 x 0.01

                    = 0.0135

conjugate base (CH3CH2COONa) molar mass = 96.06 g/mol

mass = moles x molar mass

          = 0.0135 x 96.06

          = 1.296 g

mass of conjugate base = 1.296 g

volume of acid:

moles of acid = 0.01 moles

molarity = number of moles / volume in L

volume in L = NUMBER OF MOLES / MOLARITY

= 0.01 /0.100 M

= 0.1 L

= 100 ml acid