CH3CO2H: 0.109M & 20ml
NaOH : 0.109M& 10ml
H20:10ml
Buffer Problem, when I solve it the ratio becomes the same and the pH is the same as the pka, is that right?
Yes, pH = pKa. WHY? because the log(base)/(acid) = 0
pH = pKa + log(base)/(acid)
pH = pKa + log (0.109/0.010)/(0.109/0.010)
pH = pKa + log (1)
pH = pKa + 0 = pKa.
To determine if the ratio becomes the same and the pH is equal to the pKa in a buffer problem, we need to calculate the initial concentrations of the acetic acid (CH3CO2H) and sodium hydroxide (NaOH) present in the solution, and then compare them to the Henderson-Hasselbalch equation.
1. Start by calculating the number of moles of acetic acid and sodium hydroxide:
Number of moles of CH3CO2H = concentration * volume = 0.109 M * 0.020 L = 0.00218 moles
Number of moles of NaOH = concentration * volume = 0.109 M * 0.010 L = 0.00109 moles
2. Next, consider the reaction between acetic acid and sodium hydroxide, which forms water (H2O) and sodium acetate (CH3CO2Na):
CH3CO2H + NaOH → CH3CO2Na + H2O
Since sodium hydroxide is a strong base, it will react completely with acetic acid.
The number of moles of CH3CO2H remaining after the reaction is:
Number of moles of CH3CO2H remaining = initial moles - moles reacted = 0.00218 - 0.00109 = 0.00109 moles
3. Now we can calculate the concentrations of acetic acid (CH3CO2H) and sodium acetate (CH3CO2Na) in the solution after the reaction:
Concentration of CH3CO2H = moles / volume = 0.00109 moles / 0.020 L = 0.0545 M
Concentration of CH3CO2Na = moles / volume = 0.00109 moles / 0.020 L = 0.0545 M
4. Finally, we can apply the Henderson-Hasselbalch equation to determine the pH of the solution:
pH = pKa + log([A-] / [HA])
Since the ratio of [A-] (concentration of CH3CO2Na) to [HA] (concentration of CH3CO2H) is 1:1 after the reaction, the log term becomes log(1) = 0.
Therefore, the pH of the buffer solution is equal to the pKa of acetic acid.
In summary, for a buffer solution prepared by mixing acetic acid and sodium hydroxide at equivalent concentrations, the ratio of the concentrations becomes 1:1 after the reaction, and the pH of the solution equals the pKa of the acid used.