what is the pH of a solution containing 25 ml of 0.10M acetic acid after 10 ml of 0.10M NaOH is added?
Let's call acetic acid HAc. It saves time in typing.
HAc + NaOH --> NaAc + H2O
moles HAc initially = M x L = ??
moles NaOH initially = M x L = ??
React the two (one is an acid, the other a base). The solution ends up with some salt formed, none of the NaOH remaining, and some HAc left over. This means a buffered solution since you have a weak acid and its salt (the NaAc is sodium acetate). Use the Henderson-Hasselbalch equation to solve for the pH.
Post your work if you get stuck.
what is henderson-hasselbach equation?
To determine the pH of the solution, we need to consider the reaction that occurs when acetic acid (HC2H3O2) reacts with sodium hydroxide (NaOH). This reaction is an acid-base reaction and produces water (H2O) and sodium acetate (NaC2H3O2).
HC2H3O2 + NaOH → H2O + NaC2H3O2
Initially, before mixing the solutions, there are 25 ml of 0.10M acetic acid. This means that the moles of acetic acid can be calculated using the formula:
moles = concentration (M) x volume (L)
moles of acetic acid = 0.10 M x (25 ml / 1000 ml/L)
= 0.10 moles
Similarly, there are 10 ml of 0.10M NaOH, resulting in:
moles of NaOH = 0.10 M x (10 ml / 1000 ml/L)
= 0.01 moles
Since acetic acid and NaOH react in a 1:1 ratio, the moles of acetic acid remaining after the reaction can be calculated by subtracting the moles of NaOH used from the initial moles of acetic acid:
moles of acetic acid remaining = moles of acetic acid - moles of NaOH
= 0.10 moles - 0.01 moles
= 0.09 moles
To find the concentration of acetic acid in the new solution after NaOH is added, we can use the formula:
concentration (M) = moles / volume (L)
concentration of acetic acid = 0.09 moles / [(25 ml + 10 ml) / 1000 ml/L]
= 0.09 moles / 0.035 L
= 2.57 M
Now that we know the concentration of acetic acid in the new solution, we can calculate the pOH using the equation:
pOH = -log10[OH-]
In this case, as the reaction proceeds, OH- ions are produced, which results in an increase in pOH. Since we started with a neutral solution, the initial pOH will be 7 (pOH of water).
To find the pH, we need to subtract the pOH from 14 (pH + pOH = 14):
pH = 14 - pOH
= 14 - (-log10[OH-])
At this point, we need to determine the concentration of OH- ions in the solution. Since acetic acid is a weak acid and does not dissociate completely, we can assume that the reaction consumes all the OH- ions added, making the concentration negligible.
Therefore, the pOH remains at 7, and the pH will be:
pH = 14 - 7
= 7
Hence, the pH of the solution will be 7.