What is the pH of a solution that results from diluting 0.30 mol acetic acid (CH
ID: 634140 • Letter: W
Question
What is the pH of a solution that results from diluting 0.30 mol acetic acid (CH3CO2H) and 0.20 mol sodium acetate (NaCH3CO2) with water to a volume of 1.0 L? (Ka of CH3CO2H = 1.8 x 10^-5? I know to find the molarity of each, considering the 1 L volume. I know to use the Henderson-Hasselbalch equation. My question is, how do I know which compound to put as the numerator in the equation: pH = pKa + log [NaCH3CO2]/[CH3CO2H]...I know this is the correct way to do it, but I need to know WHY sodium acetate is the numerator and not acetic acid. Thanks!
Explanation / Answer
pH = pKa + log [A?][HA]
this is because we have main reaction as AH ? H+ + A-
so in writing the disociation cnst, we take right hand side as numerator and left hand side as ddenominator
So according pH = pKa + log [NaCH3CO2]/[CH3CO2H]
NaCH3CO2 is in numerator, becausse it lies on the right hand side and other on the left hand side
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