Question

how much 5.60 M NaOH must be added to a 420.0 ml of a buffer that is 0.0205 M acetic acid and 0.0245 M sodium acetate to raise the pH to 5.75?

Answer 1

Calculation of initial moles

n CH3COOH = volume in L x molarity

=0.420 L x 0.0205 M

= 0.00861 mol CH3COOH

n CH3COONa

= 0.420 L x 0.0245 M

= 0.01029 mol

Lets moles of NaOH to be x

Now NaOH is base and it reacts with acid in 1 : 1 ratio and convert acid to base.

Lets show the reaction.

NaOH (aq) + CH3COOH (aq) --- > CH3COONa(aq) + H2O (l)

Calculation of final moles

n CH3COOH = (0.00861 -x) mol

n CH3COONa = (0.01029 +x) mol

ka of acetic acid = 1.75 E-5

pka = - log (1.75 E-5)

pka = 4.76

Lets use Henderson Hasselbalch equation

pH = pka + log ([sodium acetate]/[Acetic acid ]

5.75 – 4.76 = log ([sodium acetate]/[Acetic acid ]

log ([sodium acetate]/[Acetic acid ] = 0.993

lets take antilog of both side

([sodium acetate]/[Acetic acid ] = Antilog (0.993)

[(0.01029+x ) mol / 0.420 L ]/[(0.00681-x)/ 0.420 L ] = 9.84

0.01029+x /0.00681-x= 9.84

Now we have to find x

0.01029+x = 9.84 * (0.00861-x)

0.01029+x = 0.0847 – 9.84 x

10.84 x = 0.0744

x = 0.0744 / 10.84

=0.00687

x = 0.00687 mol NaOH

volume of NaOH = mol / molarity

= 0.00687 mol / 5.60 M

= 0.00123 L

= 1.23 mL

Volume of NaOH added = 1.23 mL