Question

Calculate how many mL of 0.100 HCl should be added to how many grams of sodium acetate dihydrate (NaOAc * 2H2O, FM 118.06) to prepare 250.0 mL of 0.100 M buffer, pH 5.00.

Answer 1

Given :

[HCl]= 0.100 M

Volume of buffer = 250.0 mL = 0.250 L

Lets calculate ratio of [NaOAc*2H2O] / [AcOH]

We know pH of this buffer is given and weak acid of this buffer is Acetic acid

We use pka value of this acid and then by using Henderson –Hasselbalch equation we get ratio.

Pka of acid = 4.75

Henderson-Hasselbalch equation

pH = pka + log ([Conjugate bas]/[Acid])

[Conjugate base ]= [Salt ]

Lets find the ratio of [NaOAc*2H2O] / [AcOH]

5.00 = 4.75 + log( [NaOAc*2H2O] / [AcOH])

log( [NaOAc*2H2O] / [AcOH]) = 0.25

Lets take antilog of both sides

( [NaOAc*2H2O] / [AcOH]) 0.25

Since volume is same we use the mole ratio

Mol of NaOAc*2H2O] / mol AcOH = 0.25

Lets calculate total moles of buffer

= Volume in L x molarity of buffer

= 0.250 L x 0.100 M

=0.0250 mol

Total moles = Mol of NaOAc*2H2O mol AcOH   = 0.0250 mol

Lets assume moles of Mol of NaOAc*2H2O = x

And moles of Mol of AcOH = y

So we have now two equations

x / y = 0.250         ….1

and x + y = 0.0250 mol ….2

Lets find the value of x in equation 1

x = 0.250 y

we plug this value in equation 2

0.250 y + y = 0.0250

Lets find y

1.250 y = 0.0250

y = 0.02 mol

x = 0.0250- y = 0.0250 – 0.02 = 0.005

moles of Mol of NaOAc*2H2O = 0.005 and moles of AcOH = 0.02 mol

Calculation of moles of HCl

From one mole of HCl , 1 mol of AcOH is produce

So number of moles of HCl = 0.02 mol

We know [HCl]= 0.02 mol / volume in L

There fore Volume in L = 0.02 mol / molarity = 0.02 mol HCl / 0.100 M

= 0.2 L

Volume of HCl in L = 200 mL

Calculation of Mass of Mol of NaOAc*2H2O

= moles of Mol of NaOAc*2H2O x F.M. of Mol of NaOAc*2H2O

= 0.005 mol x 118.06 g/mol

= 0.590 g

Mass of Mol of NaOAc*2H2O = 0.590 g