Final answer:
To calculate the pH of the solution after adding 100 ml of 0.1 M NaOH to 150 ml of 0.2 M HOAc, we must consider the buffer system formed by the reaction. We can use the Henderson-Hasselbalch equation to find the pH, which will be slightly lower than the pKa of acetic acid (4.76) due to excess acetic acid. pH will be slightly less than 4.76
Step-by-step explanation:
To determine the pH of the solution when 100 ml of 0.1 N NaOH is added to 150 ml of 0.2 M HOAc, we need to first convert the normality of NaOH to molarity. Since NaOH is a monoprotic base, its normality is equal to its molarity, making its concentration 0.1 M. Next, we will calculate the number of moles of NaOH and HOAc.
The number of moles of NaOH is given by its volume times its concentration, which is 100 ml * 0.1 M = 0.01 moles. The number of moles of HOAc is 150 ml * 0.2 M = 0.03 moles. When NaOH is added to HOAc, it will react in a 1:1 ratio to form water and the acetate ion (CH3COO-).
Since we have more HOAc than NaOH, after the reaction we will have some remaining acetic acid and some acetate ions, forming a buffer solution. We can then use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution:
pH = pKa + log([A-]/[HA])
where pKa for acetic acid is approximately 4.76.
With excess acetic acid, the concentration of acetic acid ([HA]) will be greater than that of the acetate ion ([A-]), and the pH will be slightly less than 4.76. The exact pH will depend on the resulting concentrations of [HA] and [A-] after the reaction.