Question

Your lab is supplied with 1.500 F solutions of HCL, NaOH, acetic acid, (Ka = 1.75x10^-5) sodium acetate. Prepare a 400.0mL o a 0.1000M acetate buffer: pH 4.622 using specified reagents

a) 1.000 F acetic acid and 1.000 F sodium acetate

b) 1.000 F sodium hydroxide

c) 1.000 F sodium acetate and 1.000F HCl

Please show all work!

Answer 1

We need to  prepare a 400.0mL of 0.1000M acetate buffer of pH 4.622.

Acetate buffer contains acetic acid(CH3COOH) as the acid and acetate ion(CH3COO-) as the salt

Hence total volume of buffer, Vt = 400 mL = 400mLx(1L / 1000mL) = 0.4L

[CH3COO-] = 0.1000M

pH = 4.622

Applying Henderson's equation

pH = pKa + log[salt] / [acid] = - log(1.75x10^-5) + log(0.100M / [acid])

=> 4.622 = 4.757 +log(0.100M / [acid])

=> log(0.100M / [acid]) = - 0.135

=> 0.100M / [acid] = antilog(-0.135) = 0.7328

=> [acid] = [acetic acid] = 0.1365M

Hence moles of acetic acid in the buffer = MxVt = 0.1365molL^-1x0.4L = 0.0546 mol

Hence moles of acetate ion in the buffer = MxVt = 0.1000molL^-1x0.4L = 0.0400 mol

Now we need to take the above calculated moles of acetic acid and acetate ion out of 1.000M acetic acid and 1.000 M sodium acetate.

moles of acetic acid = 0.0546 mol = MxV = 1.000molL^-1xV

=> V = 0.0546mol / 1.000molL^-1 = 0.0546L = 54.6 mL 1.000M acetic acid

moles of acetate ion = 0.0400 mol = MxV = 1.000molL^-1xV

=> V = 0.0400 mol / 1.000molL^-1 = 0.0400 L = 40.0 mL 1.000M acetate ion

Hence we need to take 54.6 mL 1.000M acetic acid and 40.0 mL 1.000M acetate ion and dilute the solution to 400 mL to prepare a  400.0mL of 0.1000M acetate buffer of pH 4.622