Chp16,Buffers

Chang, 7 th Ed  Edition, Chapter 16, Worksheet #1

S. B. Piepho, Fall 2002

Buffers

A buffer is a solution that resists changes in pH. The pH of a buffer changes very little when small amounts of a strong acid or strong base are added to the buffer. A buffer consists of approximately equal amounts of a conjugate weak acid/weak base pair in equilibrium with each other. Strong acids and their conjugate bases don’t produce a buffer since strong acid ionization is complete: there is no equilibrium!  Acidic Buffers In an acidic buffer both the weak acid and conjugate base are present initially in roughly equal concentrations. The equilibrium is

HA weak acid For a weak acid,  K a

=

[H + ][A –  ] [HA]

H+

↔

 so  p K a

+

A– conjugate base

  [ A− ]     . Thus  pH = p K  + = pH − log  [HA]  a

  [A − ]     : log  [HA]  

 Henderson-Hasselbalch lbalch equation equation . this is the Henderson-Hasse

These equations are useful for estimating the pH of a buffer. When [A -] = [HA], it follows that Ka = [H+] and pH = pK a. Acidic Acidi c Buffer Buffe r Example: Exampl e: Acetic acid (HC 2H3O2), sodi s odium um acet a cetate ate (NaC 2H3O2) + NaC2H3O2 is a soluble salt that dissolves in water to give Na  and C2H3O2- ions. Na+ ions are neutral spectator ions, so can be ignored. The C 2H3O2- ions provide the conjugate base for the buffer. The buffer equilibrium is HC2H3O2 acetic acid

↔

H+

+

C2H3O2– acetate ion

For acetic acid, K a(HC2H3O2) = 1.8 x 10-5 so pK a = -log10K a = 4.74. Thus when [HC 2H3O2] = [C2H3O2–], pH = pK a = 4.74.  Basic Buffers In a basic buffer both the weak base and conjugate acid are present initially in roughly equal concentrations. The equilibrium is

B + weak base

H2O

↔

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BH+

+

OH–

Chang, 7 th Edition, Chapter 16, Worksheet #1

For a weak base,  K b  p K b

=

 [BH + ][OH –  ] [B]

S. B. Piepho, Fall 2002

 so  p K b

  [BH + ]     = = pOH − log   [B]  

    [BH+ ]   [BH + ]       . Since pK a = 14.00 = (14.00 − pH) − log . Thus  pH = 14.00 − p K  − log   [B]     [B]   b

  [B]    [BH + ] . This is once again the  Henderson-Hasselbalch equation .

– pK b,  pH = p K a (BH ) + log +

These equations are useful for estimating the pH of a buffer. When [BH +] = [B], K b = [OH–]. Consequently, pOH = p K b for the buffer. It follows that pH = 14.00 – p K b = pK a(BH+) for the buffer. Basic Buffer Example: Ammonia (NH 3), ammonium chloride ( NH 4Cl) NH4Cl is a soluble salt that dissolves in water to give NH 4+ and Cl– ions. Cl– ions are neutral spectator ions, so can be ignored. The NH 4+ ions provide the conjugate acid for the buffer. The buffer equilibrium is NH3 + ammonia

H2O

↔

OH–

+ NH4+ ammonium ion

For ammonia, K b(NH3) = 1.8 x 10-5 so pK b = -log10K b = 4.74. Thus when [NH 3] = [NH4+], pOH = 4.74 and pH = pK a(NH4+) = 14.00 – 4.74 = 9.26.

 Neutral Buffers Neutral buffers have a pH close to 7.00. A good example is a NaH 2PO4 /Na2HPO4 buffer. Since Na+ ions are neutral spectator ions, this is a dihydrogen phosphate/hydrogen phosphate (H2PO4– /HPO42–) buffer. The buffer equilibrium is ↔ H2PO4– dihydrogen phosphate ion

H+

+

HPO42– hydrogen phosphate ion

2– 

Here  K a =

 [H + ][HPO 4 ] −

[H2PO 4 ]

= 6.2 x 10–8. If [H2PO4-] = [HPO42-], K a = [H+] and pH = pK a = 7.21.

 Response of a Buffer to the Addition of a Strong Acid or a Strong Base Added Acid. When a small amount of a strong acid such as HCl is added to a buffer, the H + from the acid reacts with the basic part  of the buffer to give more of the acidic part  of the buffer. The reaction is assumed to go 100%. The new concentration of the acidic part  of the buffer (increased from the initial value) and the new concentration of the basic part  of the buffer (decreased from the initial value) are then used to calculate the pH of the buffer.

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Chang, 7 th Edition, Chapter 16, Worksheet #1

S. B. Piepho, Fall 2002

Added Base. When a small amount of a strong base is added to a buffer, the OH – reacts with the acidic part  of the buffer to give more of the basic part  of the buffer. The reaction is assumed to go 100%. The new concentration of the acidic part  of the buffer (decreased from the initial value) and the new concentration of the basic part  of the buffer (increased from the initial value) are then used to calculate the pH of the buffer.

Exercises 1. Identify which  of the following mixed systems could function as a buffer   solution. For each system that can function as a buffer, write the equilibrium equation for the conjugate acid/base  pair in the buffer system (a) KF/HF (b) NH3 /NH4Br (c)KNO3 /HNO3 (d) Na2CO3 /NaHCO3

 Answer : (a) Buffer: HF ↔ H+ + F– (b) Buffer: NH 3 + H2O ↔ NH4+ + OH– (or NH4+ ↔ H+ + NH3) (c) Not a buffer since HNO 3 is a strong acid. It is 100% ionized. (d) Buffer: HCO3– ↔ H+ + CO32–

2. What is the pH of a 1 L solution containing 0.240 mol HC 2H3O2 and 0.180 mol NaC 2H3O2? K a(HC2H3O2) = 1.8 x 10-5. HINT: Begin by filling out the equilibrium table below. Balanced Equation HC 2H3O2 H+ + C2H3O2– Initial Concentration (M) Change (M) Equilibrium Concentration (M)

 Answer :

Balanced Equation Initial Concentration (M) Change (M) Equilibrium Concentration (M)

0.240 -x 0.240 - x

−

 K a =

 [H + ][C2 H3 O2 ] [HC2 H3 O2 ]

H+

HC2H3O2

−5

= 1.8 × 10

=

x (0.180 + x )

≈

(0.240 -  x )

0 x x

x (0.180)

0.240

ionization is < 5%; it is in this case. Thus  x = 2.4 x 10-5 = [H+]. pH = 4.62.

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+

C2H3O2– 0.180 x 0.180 + x

. This approximation is OK if the %

Chang, 7 th Edition, Chapter 16, Worksheet #1

S. B. Piepho, Fall 2002

3. What would be the pH of the buffer from #2 above if 0.010 mol of HCl were added per liter of  the buffer?

 Answer : The new initial concentrations are: [C2H3O2-] = (0.180 - 0.010) M = 0.170 M (basic part decreases) [HC2H3O2] = (0.240 + 0.010) M = 0.250 M (acidic part increases) −

Thus  K a =

 [H + ][C2 H3 O2 ] [HC2 H3 O2 ]

−5

= 1.8 × 10

=

x (0.170 + x )

≈

(0.250 - x )

x (0.170)

0.250

. Thus x = [H+] = 2.647 x 10 -5; pH =

4.58. Note that the pH has decreased very little!!

4. What would be the pH of the buffer from #2 above if 0.010 mol NaOH were added per liter of the buffer?

 Answer: The new initial concentrations are: [C2H3O2-] = (0.180 + 0.010) M = 0.190 M (basic part increases) [HC2H3O2] = (0.240 - .010) M = 0.230 M (acidic part decreases) −

Thus  K a =

 [H + ][C2 H3 O2 ] [HC2 H3 O2 ]

= 1.8 ×10

−5

=

x (0.190 + x )

0.230 - x

≈

x (0.190)

0.230

. Thus x = [H+] = 2.179 x 10-5; pH

= 4.66. Note that the pH has increased very little!!

5. What would be the pH of 0.010 M HCl without the buffer?

 Answer: HCl is a strong acid and is therefore 100% ionized. Thus [H +] = 0.010M, so the pH = 2.00. Note how much lower the pH is than it was with the buffer in #3 above (pH = 2.00 here versus 4.58 in #3 above)!

6. What would be the pH of 0.010 M NaOH without the buffer?

 Answer:

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Chang, 7 th Edition, Chapter 16, Worksheet #1

S. B. Piepho, Fall 2002

NaOH is a strong base and is therefore 100% ionized. Thus [OH –] = 0.010M, so the pOH = 2.00. Thus pH = 14.00 – pOH = 12.00. Note how much higher the pH is than it was with the buffer in #4 above (pH = 12.00 here versus 4.66 in #4 above)!

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