Question

Which equation can be used to find the pH of a buffer. Then calculate the pH of a buffer containing 0.20 M CH3COOH and 0.20 M CH3COONa. What is the pH after adding 10.0 mL of 0.10 M HCl to 65.0 mL of the buffer?

Answer 1

Given

[CH3COOH]= 0.20 M

[CH3COONa]= 0.20 M

Since this system is buffer and we have concentration of both conjugate base and acid so we can use Henderson Hasselbalch equation

pH = pka + log ([conj base ]/[acid ])

pka = - log ka = - log 1.8 E-5 = 4.74

volume of buffer = 65.0 mL = 0.065 L

Calculation of moles of acid and its conj base

Moles of CH3COOH = moles of CH3COONa = Molarity x volume inL

= 0.20 M x 0.065 L

= 0.013 mol

Mole of HCl = Volume in L x molarity

= 0.010 L x 0.10 M = 0.001 mol

Total volume = 0.065 + 0.010 )L = 0.075 L

Les find equilibrium concentrations of acid and base

HCl reacts with base and form CH3COOH

Lets show the reaction

HCl (aq) + CH3COO- (aq) - -- > CH3COOH (aq) + Cl- (aq)

So this decrease one mole of base and form one mole of acid

So concentration of both at equilibrium

[CH3COOH]= (Initial moles + moles of HCl ) / 0.075 L

= 0.013 mol + 0.001 mol /0.075 L = 0.014 mol /0.075 L

= 0.1867 M

[CH3COO-]= (0.013-0.001) mol/ 0.075 L

= 0.16 M

Now we use Henderson equation

pH = 4.74 + log ( 0.16 M/ 0.1867 M )

= 4.67

pH of buffer = 4.67