Question

Calculate how many milliliters of 0.1 M HCL should be added to how many grams of sodium acetate dihydrate (NaOAc*2H20, FM 118.06) to prepare 250 mL of 0.1 M buffer, pH=5.0

Answer 1

moles of buffer = 250 x 0.1 /1000= 0.025

acid + salt = 0.025 --------------------->1

sodium acetate + HCl ---------------------> acetic acid

x mol                      0.1 V mol                      0           ---------------------> initial

x -0.1V                    0                                0.1 V --------------------------> after reaction

pH = pKa + log [salt/acid]

5.0 = 4.74 + log (salt/acid)

0.26 = log (salt/acid)

1.82 = salt/acid

salt = 1.82 acid -----------------> 2

acid + 1.82 acid = 0.025

acid moles = 0.0089

salt moles = 0.016

acid moles = 0.0089

molarity x volume = 0.0089

0.1 x volume = 0.0089

volume = 0.089 L

volume of HCl = 89 ml

salt moles = 0.016

mass /molar mass = 0.016

mass = 0.016 x molar mass = 0.016 x 118.06 = 1.89 g

sodium acetate dihydrate mass = 1.89 g