calculate the pH of a solution that results from mixing 150 mL of .25 CH3COOH (aq) with 225.0 mL of .30 M CH3COONa (aq). (Ka (CH3COOH)=1.8 *10^-5)
- anonymousLv 72 years ago
number of moles of CH₃COOH :
(0.150 L) x (0.25 moles CH₃COOH / 1 L) = 0.0375 moles CH₃COOH
number of moles of CH₃COO⁻ :
(0.2250 L) x (0.30 moles CH₃COO⁻ / 1 L) = 0.0675 moles CH₃COO⁻
Note: Because [A⁻]/[HA] in the Henderson-Hasselbalch equation is a ratio of molarities, it is also a ratio of moles. So you can either calculate the molarities of A⁻ and HA, or you can just use the moles of each that you have calculated (which is what I will do).
pH = pKa + log([A⁻]/[HA])
= -log(1.8 x 10⁻⁵) + log(0.0675 / 0.0375)
= 5.00 …….. to 2 sig figs
Another comment I found on the internet about moles vs molarity in the H-H equation:
In the Henderson-Hasselbalch equation, you SHOULD put in the molarity, but since the total volume is the same for both the base and the acid (they are sitting in the same solution, so they have the same total volume of solution), the volumes will cancel. For example say the total volume is 10 L, [(moles base)/(10)]/[moles acid)/(10)] = (moles base)/(moles acid). If this makes you uncomfortable, or you are afraid you may not divide by the volume when you have to (for instance at the stoichiometric point), then by all means calculate the molarity and plug that into the Henderson-Hasselbalch equation.