Question

A. Calculate molarity of glacial acetic acid( mol. weight=60g/mol, specific density 1.049 g/mL)

B.Calculate the amount (in grams) of sodium acetate (Mol. weight= 82g/mole) and the amount (in mL) of glacial acetic acid required to make 1 L of 0.1 M sodium acetate buffer at pH 4.5. The pKa of acetic acid is 4.8.

Answer 1

A]

Molarity = Moles /Volume in L = mass*1000 / Molarmass*Volume in ml

Information is not provided to calculate molarity

So , I took 5 ml of glacial acetic acid

d = mass /volume

1.049 = mass / 5

mass = 5.245 gms of glacial acetic acid

Molarity = 5.245 *1000 / 60*5 = 17.4833 M

B]

ACidi buffer

pH = pKa + log [Sodium acetate ] / [Acetic acid]

4.5 = 4.8 + log [Sodium acetate ] / [Acetic acid ]

Sodium acetate / Acetic acid = 0.5012

x / y = 0.5012

Moles of Sodium acetate = x

Moles of Acetic acid = y

Moles of buffer = Molarity*V in L = 1*0.1 = 0.1

x + y = 0.1 (Moles of buffer = moles of both components )

x / y = 0.5012

y = 0.0666

x = 0.0334

Mass of acetic acid = y*Molar mass = 2 gms

d = m / v

v = m /d = 2 / 1.049 = 1.9 ml of acetic acid

Mass of sodium acetate = x*molar mass = 5.4612 gms