Question

A buffer is prepared by mixing 4.00g of sodium acetate (NaC₂H₃O₂.3H₂O) and 10.00mL 3.0M acetic acid. K_a = 1.8x10^-5 for acetic acid.

a) Calculate the pH of the buffer.

b) If we add 5.00mL of 1.0M HCl to 25mL of the buffer, what is the new pH of the buffer?

SHOW STEPS

Answer 1

m = 4g NaAc

V = 10 ml

M = 3 M HAc

Ka = 1.8*10^-5

a) find pH

this is a buffer so apply buffer law

pH = pK a+ log(A-/HA)

pka = -log(Ka) = -log(1.8*10^-5) = 4.75

we need to find concnetratino of 4 g of NAAc in 10 ml

MW of NaAc = 82.0343

mol NAAc = 4/82.0343 = 0.04876 mol

M = mol/V = 0.04876/0.01 = 4.876 M

Subtitute in pH equation

pH = pK a+ log(A-/HA) = 4.75 + log(4.876/3) = 4.96

b)

if adding V = 50 ml of M = 1 HCl to 25 ml of buffer... what will be the new pH

A- will remain with the same concnetration

HA will change, will increase

[A-] = 4.96 since concnetration is maintained

find moles of HCl added

mol = M*V = 1*5= 5 mmol of HCl

Final concnetration

M = mol/VT = 5 / (5+25) = 0.1667 M of HCl

HA <-> H+ and A-

Since H+ are added in the form of HCl (HCl--> H+ and Cl-)

the equilibrium will be shifted

H+ and A-<-> HA

Ka = [H+][A-]/[HA]

[H+] = x

[A-] = 4.96 - 0.1667 = 4.7933

[HA] = 3 + 0.1667 = 3.1667

Now apply buffer equation

pH = pKa+ log(A-/HA) = 4.75 + log( 4.7933/3.1667) = 4.93