.01 mols of a weak acid is dissolved in .1 L of water. That solution is then reacted with .01 mols of NaOH. Would the resulting solution be acidic, basic, or neutral? Why? What reaction equations support the explanation?
I think the answer is that the resulting solution would be basic because NaOH is strong and the acid is weak but I'm not sure what reaction equation explains that.
It isn't a matter of strong base vs weak acid as much as it is the number of mols of each and how the salt reacts with H2O.
HA + NaOH ==> H2O + NaA
mols HA = 0.01
mols NaOH = 0.01
So the solution EXACTLY neutralizes and there is no excess NaOH or HA. You might think the solution would therefore be neutral and it would be if it were a strong acid neutralizing a strong base. You are absolutely right; the solution will be basic. The reason is because the anion of the weak acid hydrolyzes in H2O as
A^- + HOH --> HA + OH^-
and the excess OH^- ions makes the solution slightly basic.
For example, a solution of acetic acid and HCl gives sodium acetate. When acetic acid and HCl are mixed in exact molar quantities, the salt produced is sodium acetate and the pH is not 7.0 but about 8.3 or so.
To determine whether the resulting solution is acidic, basic, or neutral, we need to consider the reaction between the weak acid and NaOH. Let's break down the steps to find the answer:
Step 1: Identify the weak acid and its dissociation equation.
Since we know that we have a weak acid, we can assume it will partially dissociate in water. Let's represent the weak acid as HA. The dissociation equation for a weak acid can be written as:
HA(aq) ⇌ H+(aq) + A-(aq)
Step 2: Calculate the concentration of the weak acid.
Given that we have 0.01 moles of the weak acid and a total volume of 0.1 L, we can determine the concentration. Concentration (C) is calculated by dividing the number of moles (n) by the volume (V):
C = n/V
C = 0.01 mol / 0.1 L
C = 0.1 M
Step 3: Determine the dissociation extent.
Since the acid is weak, it will not completely dissociate. That means only a fraction of the weak acid will break apart into H+ and A- ions. Let's denote the extent of dissociation as 'x'. Thus, the concentration of H+ ions would be 'x' (in M), and the concentration of the undissociated acid would be (0.1 - x) M.
Step 4: Analyze the neutralization reaction with NaOH.
For every mole of NaOH reacting with the weak acid, one mole of OH- ions will react with one mole of H+ ions. The reaction equation is:
H+(aq) + OH-(aq) → H2O(l)
Since the number of moles of NaOH added is equal to the number of moles of H+ ions, it will consume all of the H+ ions if the reaction goes to completion. However, in this case, the reaction will not proceed to completion because the concentration of the weak acid is much higher than the NaOH concentration.
Step 5: Determine the resulting solution's pH.
To determine the pH of the solution after the reaction, we need to calculate the concentration of OH- ions that will be present in excess after the reaction.
With 0.01 moles of NaOH, the concentration of OH- ions would be:
OH- concentration = (0.01 mol) / (0.1 L) = 0.1 M
Since OH- and H+ ions react in a 1:1 ratio, the remaining OH- ions must equal the excess concentration of H+ ions after the reaction. Therefore, the excess OH- concentration is 0.1 M.
Step 6: Determine if the resulting solution is acidic, basic, or neutral.
The resulting solution will be basic because the concentration of OH- ions in the solution is higher than the concentration of H+ ions. The excess OH- ions make the solution more alkaline.
Reaction equations to support the explanation:
HA(aq) ⇌ H+(aq) + A-(aq) (Dissociation of weak acid)
H+(aq) + OH-(aq) → H2O(l) (Neutralization reaction between H+ and OH- ions)
Remember that this explanation assumes the weak acid does not undergo any other reactions, and the solution is not strongly buffered.