Question

One liter of a 0.1M Tris buffer (pKa=8.3) is adjusted to a pH of 2.0. A) What are the concentrations of the conjugate base and weak acid at this pH? B) What is the pH when 1.5mL of 3.0M HCl is added to this buffer? Is Tris a good buffer at this pH? Why? C) What is the pH when 1.5mL of 3.0M NaOH is added to this buffer?

Answer 1

Part.A :-

Given concentration of Tris buffer i.e [ Tris buffer ] = 0.1 M

also [ Tris buffer ] = [ Conjugate base ] + [ Acid ]

0.1 M = [ Conjugate base ] + [ Acid ]  

[ Acid ] = 0.1 M -  [ Conjugate base ] .............(1)

Now from Henderson-Hasselbalch equation , we have

pH = pK_a + log [ Conjugate base ] / [ Acid ]

2.0 = 8.3 +  log [ Conjugate base ] / [ Acid ]

log [ Conjugate base ] / [ Acid ] = 2.0 - 8.3

log [ Conjugate base ] / [ Acid ] = - 6.3

[ Conjugate base ] / [ Acid ] = 10^-6.3

[ Conjugate base ] / [ Acid ] = 5.01 x 10^-7

[ Conjugate base ] = 5.01 x 10^-7 x [ Acid ]   ..............(2)

Now put the value of equation (2) in equation (1) , we have

[ Acid ] = 0.1 M -  5.01 x 10^-7

[ Acid ] = 0.1 M ( approx.)

and from equation (2)

[ Conjugate base ] = 5.01 x 10^-7 x 0.1 M

= 5.01 x 10^-8 M

Hence  [ Acid ] = 0.1 M and [ Conjugate base ] = 5.01 x 10^-8 M

Part. B :-

Number of moles of HCl = 3.0 M x 0.00152 L

= 0.0045 moles

Number of moles Conjugate base in 1 L of buffer = 5.01 x 10^-8 M x 1 L = 5.01 x 10^-8 moles  

Number of moles of weak acid = 0.1 M x 1 L = 0.1 moles

Now all conjugate base is neutralized by HCl

therefore

Molarity of HCl in solution = 0.0045 moles / 1 L = 0.0045 M

i.e [ H⁺ ] in HCl = 0.0045 M

also pH = 2.0

therefore [ H⁺ ] in buffer = 10^-2 M = 0.01 M ( because pH = - log [ H⁺] and [ H⁺ ] = 10^-pH )

Now Total concentration of H⁺ i.e [H⁺]_Total = 0.0045 + 0.01 = 0.0145 M

pH = - log 0.0145

pH = - ( - 1.84 )

pH = 1.84

Hence pH = 1.84

Part. C :-

Number of moles of NaOH = 3.0 M x 0.00152 L

= 0.0045 moles

Number of moles Conjugate base ( say A^-) in 1 L of buffer = 5.01 x 10^-8 M x 1 L = 5.01 x 10^-8 moles  

Number of moles of weak acid ( say HA ) = 0.1 M x 1 L = 0.1 moles

Now ICF table is :

HA + OH^- ----------------> A^- + H₂O

I 0.1 mol 0.0045 mol 5.01 x 10^-8 mol

C - 0.0045 mol - 0.0045 mol + 0.0045 mol

F 0.0955 mol 0 mol 0.0045 mol ( approx. )

Now again from Henderson-Hasselbalch equation , we have

pH = 8.3 + log 0.0045 / 0.0955

pH = 8.3 + log 0.04712

pH = 8.3 - 1.3268

pH = 6.97

Hence pH = 6.97