Chemistry Equilibrium MDCAT 2015
PMDC Verified Question 93 of 102
What is the correct relation between pH and pKa?
A
\( \text{pH} = \text{pKa} + \log \left( \frac{[\text{Acid}]}{[\text{Base}]} \right) \)
B
\( \text{pH} = \text{pKa} - \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \)
C
\( \text{pH} = \text{pKa} - \log \left( \frac{[\text{Acid}]}{[\text{Base}]} \right) \)
D
\( \text{pKa} = \text{pH} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: \( \text{pH} = \text{pKa} - \log \left( \frac{[\text{Acid}]}{[\text{Base}]} \right) \)
Concept:

The Henderson-Hasselbalch equation defines the relationship between the pH of a buffer, the \( \text{pKa} \) of the weak acid, and the concentrations of the acid and its conjugate base.

Formula:

$$ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) $$

Solution:

  • The standard form of the equation has a positive logarithm of Base over Acid.


  • By logarithmic properties, \( +\log(x/y) = -\log(y/x) \).


  • Applying this: \( +\log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) = -\log \left( \frac{[\text{Acid}]}{[\text{Base}]} \right) \).


  • Substituting this back gives \( \text{pH} = \text{pK}_a - \log \left( \frac{[\text{Acid}]}{[\text{Base}]} \right) \).


Why other options are incorrect:

Option A has a positive sign but inverted ratio. Option B applies a negative sign incorrectly to the standard ratio. Option D algebraically fails when rearranging the standard formula.

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