Intro to the problem: To understand how buffers use reserves of conjugate acid a

Question

Intro to the problem:

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, CH3COOH, and its conjugate base, the acetate ion CH3COO. Because ions cannot simply be added to a solution, the conjugate base is added in a salt form (e.g., sodium acetate NaCH3COO).

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++CH3COOCH3COOH

Similarly, any added OH ions will be neutralized by a reaction with the conjugate acid:

OH+CH3COOHCH3COO+H2O

This buffer system is described by the Henderson-Hasselbalch equation

pH=pKa+log[conjugate base][conjugate acid]    

Part A

A beaker with 1.80×102 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 5.60 mL of a 0.330 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Express your answer numerically to two decimal places. Use a minus ( ) sign if the pH has decreased.

Explanation / Answer

we know that

for buffers

pH = pKa + log [salt / acid ]

in this case

pH = pKa + log [ CH3COO- / CH3COOH ]

5 = 4.74 + log [ CH3COO- / CH3COOH ]

[ CH3COO- / CH3COOH ] = 1.82

[CH3COO-] = 1.82 [CH3COOH]

now

given

total molarity = 0.1

so

[CH3COO-] + [CH3COOH] = 0.1

1.82 [CH3COOH] + [CH3COOH] = 0.1

[CH3COOH] = 0.03546

[CH3COO-] = 1.82 x 0.035461 = 0.06454

now

we know that

moles = molarity x volume (L)

so

moles of HCL added = 0.33 x 5.6 x 10-3 = 1.848 x 10-3

moles of CH3COOH = 0.03546 x 180 x 10-3 = 6.3828 x 10-3

moles of Ch3COO- = 0.06456 x 180 x 10-3 = 11.6208 x 10-3

now

the reaction is

CH3COO- + H+ --> CH3COOH

we can see that

moles of CH3COO- reacted = moles of H+ added = 1.848 x 10-3

moles of CH3COOH formed = moles of H+ added = 1.848 x 10-3

now

finally

moles of CH3COOH = 6.3828 x 10-3 + 1.848 x 10-3 = 8.2308 x 10-3

moles of CH3COO- = 11.6208 x 10-3 - 1.848 x 10-3 = 9.7728 x 10-3

now

pH = pKa + log [ CH3COO- / CH3COOH]

pH = 4.74 + log [ 9.7728 x 10-3 / 8.2308 x 10-3 ]

pH = 4.8145

so

the pH of the solution is 4.8145

pH change = 4.8145 - 5

pH change = - 0.1854

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