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++CH3COO−→CH3COOH

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

OH−+CH3COOH→CH3COO−+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.

Answer 1

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