The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.220 moles of a monoprotic weak acid (Ka = 3.7 × 10-5) is titrated with NaOH, what is the pH of the solution at the half-equivalence point?
Well, well, well, looks like we've got a titration on our hands! But don't worry, I'm here to bring some levity to the equation (pun intended).
At the half-equivalence point, we know that half of the analyte has reacted to form its conjugate, and the other half is still hangin' out, mindin' its own business. So, let's put our comedic chemistry cap on and figure out the pH at this point.
Now, since we're dealing with a weak acid here, we'll have to consider the equilibrium between the acid and its conjugate base. The expression for the acid dissociation, Ka, is given as [H+][A-]/[HA]. But at the half-equivalence point, half of our weak acid HA has reacted, meaning the concentration of [HA] will be halved.
So, we can rewrite the equilibrium expression as [H+][A-]/[HA/2]. Since the concentration of [A-] is equal to [H+], because it's this lil' rascally half that reacted, we can simplify the expression to [H+]^2/[HA/2]. Phew, that was a lot of math talk!
Now, we know the concentration of [HA] at the half-equivalence point is half of what it was at the start, which means it'll be 0.220/2 = 0.110 M.
Solving for [H+], we can rearrange the equation: [H+]^2 = Ka * [HA/2]. Plugging in the values, we get [H+]^2 = (3.7 × 10-5) * (0.110/2).
Now, to find the pH, we need to take the negative logarithm of [H+]. So, pH = -log([H+]). Just plug in the value of [H+] we calculated, and voila!
Remember, I'm just a little clown bot, so I can't do math in real-time, but I promise it's doable. Now go put your chemistry skills to work and solve that equation!
To find the pH of the solution at the half-equivalence point of the titration, we need to determine the concentration of the acid and its conjugate base.
Given:
- Moles of monoprotic weak acid (HA): 0.220 moles
- Ka (acid dissociation constant) of the weak acid (HA): 3.7 × 10^(-5)
Since the half-equivalence point occurs when half of the acid has reacted, we can calculate the moles of the weak acid and its conjugate base at the half-equivalence point.
Initially, we have 0.220 moles of acid. At the half-equivalence point, half of this will have reacted, leaving us with 0.220/2 = 0.110 moles of acid remaining.
Since the weak acid (HA) dissociates to form H+ and its conjugate base (A-), the number of moles of acid remaining at the half-equivalence point is equal to the number of moles of the conjugate base.
Next, we can calculate the concentrations of acid (HA) and conjugate base (A-) at the half-equivalence point using the volume of the solution.
Note: The volume of the solution is not given in the question. Without the volume information, we cannot directly calculate the concentrations and pH at the half-equivalence point.
If you have the volume information, please provide it, and I will be able to calculate the pH at the half-equivalence point for you.
To determine the pH at the half-equivalence point of a weak acid titration, we need to consider the reaction that takes place during the titration.
In this case, the monoprotic weak acid is reacting with NaOH, which is a strong base. The reaction can be represented as follows:
HA + OH- ⇌ A- + H2O
At the half-equivalence point, half of the moles of the weak acid have reacted to form its conjugate base. Since the weak acid is monoprotic, this means that 0.220/2 = 0.110 moles of the weak acid have reacted.
Now, we can calculate the concentration of the weak acid and its conjugate base at the half-equivalence point. Since the volume of the titration is not given, we'll assume it to be 1 liter for simplicity.
The initial concentration of the weak acid is given by:
[HA] = (moles of HA) / (volume in liters)
= 0.220 mol / 1 L
= 0.220 M
At the half-equivalence point, the concentration of the weak acid is halved:
[HA] = 0.220 M / 2
= 0.110 M
Since the weak acid has dissociated into its conjugate base, its concentration is now equal to the concentration of the weak acid:
[A-] = 0.110 M
Now, we can use the Ka expression for the weak acid to calculate the pH. The Ka expression is given by:
Ka = ([A-] * [H+]) / [HA]
Since we are at the half-equivalence point, the concentration of the weak acid is equal to the concentration of the conjugate base:
Ka = ([A-] * [H+]) / ([A-])
Substituting the known values and solving for [H+], we get:
Ka = [H+]
Ka = 3.7 × 10^-5
[H+]^2 = 3.7 × 10^-5
[H+] = √(3.7 × 10^-5)
[H+] = 6.08 × 10^-3
To determine the pH, we take the negative logarithm (base 10) of the [H+] concentration:
pH = -log([H+])
= -log(6.08 × 10^-3)
= -(-2.213)
= 2.213
Therefore, the pH of the solution at the half-equivalence point is 2.213.
At the half point, pH = pKa. You can work it out if necessary.
pH = pKa + log(base)/(acid)
At the half way point, as you've explained above, (base) formed = (acid) left which makes (base) = (acid)
So base/acid = 1, log 1 = 0 and pH = pKa.