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Answer :
Question: In a solution of 168.3 mM acetate buffer (this is the concentration of all ionization states of the acetate/acetic acid combined), what is the pH of the solution if the concentration of acetate is 36.6 mM (just the deprotonated state)? The pKa of acetic acid is 4.76. Report your answer to two decimal places.
To find the pH of the solution, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the concentration of the acid and its conjugate base. The equation is:
pH = pKa + log([A-]/[HA])
Where:
pH is the pH of the solution
pKa is the acid dissociation constant
[A-] is the concentration of the acetate ion (deprotonated state)
[HA] is the concentration of the acetic acid
In this case, the concentration of acetate ([A-]) is given as 36.6 mM, and the pKa of acetic acid is 4.76.
Now let's plug these values into the equation:
pH = 4.76 + log(36.6/([HA]))
Since we know the concentration of acetate ([A-]), we can subtract it from the total concentration of the buffer to find the concentration of acetic acid ([HA]):
[HA] = 168.3 - 36.6 = 131.7 mM
Now we can substitute this value into the equation:
pH = 4.76 + log(36.6/131.7)
Calculating this expression, we get:
pH ≈ 4.76 + log(0.2779)
Using a calculator, we find:
pH ≈ 4.76 + (-0.5562)
pH ≈ 4.20
Therefore, the pH of the solution is approximately 4.20 when the concentration of acetate is 36.6 mM.
Know more about Henderson-Hasselbalch equation here:
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