To understand how buffers use reserves of conjugate acid and conjugate base to c

Question

To understand how buffers use reserves of conjugate acid and conjugate base to counteract the effects of acid or base addition on pH. A buffer is a mixture of a conjugate acid-base pair. In other words, it is a solution that contains a weak acid and its conjugate base, or a weak base and its conjugate acid. For example, an acetic acid buffer consists of acetic acid. CH_3 COOH. and its conjugate base, the acetate ion CH_3 COO^-. Because ions cannot simply be added to a solution, the conjugate base is added in a salt form (e.g.. sodium acetate NaCH_3 COO). Buffers work because the conjugate acid-base pair work together to neutralize the addition of H^+ or OH^- ions. Thus, for example, if H' ions are added to the acetate buffer described above, they will be largely removed from solution by the reaction of H^+ with the conjugate base: H^+ + CH_3 COO^- rightarrow CH_3 COOH Similarly, any added OH^- ions will be neutralized by a reaction with the conjugate acid: OH^- + CH_3 COOH rightarrow CH_3 COO^- + H_2 O This buffer system is described by the Henderson-Hasselbalch equation A beaker with 1.10 times 10^2 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.70 mL of a 0.440 MHCl solution to the beaker. How much will the pH change? The pK_a of acetic acid is 4.740.

Explanation / Answer

Volume of buffer= 110ml=0.11L

Let A-= conjugate base and HA= acetic acid

Hence [A-]+[HA]=0.1 (1)

Given pH= 5 and PKa= 4.74,

pH= pKa+ log{[A-]/[HA]}

5= 4.74+ log[A-]/[HA]

[A-]/[HA] = 1.82

[A-]= 1.82[HA], from Eq.1, [HA]+1.82[HA]= 0.1, [HA]= 0.1/2.82= 0.035 , [A-]= 1.82*0.035=0.065

Moles of [HA] = Molarity* Volume(L)= 0.035*0.11=0.00385, moles of A- = 0.065*0.11= 0.00715

Moles of HCl added = 0.44*8.7/1000 =0.0038

The reaction between CH3COONa ( from conjugate base A-) and HCl is

CH3COONa+ HCl ----àCH3COOH+ NaCl

Molar ratio of CH3COONa: HCl =1:1

Here HCl is limiting and moles of CH3COOH formed additionally= 0.0038

Hence [HA] =0.0038+0.00385= 0.00765, [A-] remaining after reaction =0.00715-0.0038 =0.00335

Volume after mixing is immaterial since molar ratio of A – to HA is same as concentration ratio

Hence pH= 4.74+log ( 0.00335/0.00765)=4.38

Change in pH= 5-4.38=0.62 , pH is decreased by 0.62

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