Consider the titration of 25.0mL of 0.10M HAc with 0.10M NaOH. That is, NaOH is added to HAc. (a)pH at the beginning of titration. (b)pH at the equivalence point of the titration. (c) pH at the midpoint of the titration.
a.) pH=-log[H+]=-log(0.10M HAc)
b.) At the 1/2 eq. pt., 1/2 of the HAc will react with NaOH. So, 25 z 10^-3L (0.01M)= moles of HAc
1/2*moles of HAc= moles of unreacted HAc
moles of unreacted HAc/total volume (37.5mL)= M of unreacted HAc
pH=-log[H+]=-log[M of unreacted HAc]
c.) Neutralization reaction, pH= 7
I apologize,
HAc + H20---> H30+ Ac
HAc= acetic acid, which is a weak acid.
a.)
Ka=1.8 x 10–5=[x][x]/0.100M-x
5% rule allows you to ignore -x
Ka=1.8 x 10–5=x^2/0.100M
solving for x,
x=1.34 x 10^-3
pH=-log(1.34 x 10^-3 M)
B.)
pKa=-log(Ka)
pH=pKa+log([Ac/HAc])
Concentrations of Ac and HAc are eqaul, pH=pKA.
C.)
Ac- + H2O ---> OH +HAc
Kb=Kw/Ka=[OH][HAc]/[Ac-]
Ac-=1/2*(0.10M)
sqrt*(Kb[Ac-])=OH
-log[OH]=pOH
14-pOH=pH
I apologize about that one; I assumed HAc was a strong acid. Oppps.
To determine the pH at different stages of the titration, we need to consider the acid-base reactions and the resulting concentrations of the species involved. Here's how we can calculate the pH at each stage:
(a) pH at the beginning of titration:
At the beginning of the titration, only HAc is present in the solution. HAc is a weak acid, so we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log ([A-]/[HA])
The pKa value for acetic acid (HAc) is around 4.76. Assuming complete ionization of HAc, we have:
[HAc] = 0.10 M
[A-] = 0 M (no NaOH has been added yet)
Therefore, the pH at the beginning of titration is approximately equal to the pKa value, which is 4.76.
(b) pH at the equivalence point:
At the equivalence point, the moles of NaOH added is stoichiometrically equal to the moles of HAc initially present. This means that all the HAc has been neutralized, resulting in a solution of sodium acetate (NaAc).
Since NaAc is a salt, it completely dissociates in water to produce Na+ and Ac- ions. The resulting solution is a sodium acetate buffer. The pH of a sodium acetate buffer can be calculated using the Henderson-Hasselbalch equation as well.
[HAc] = [A-] (at the equivalence point)
Let's assume that the volume of added NaOH until the equivalence point is Veq mL.
[HAc] = (0.10 M * 25 mL) / (25 mL + Veq mL)
[A-] = (0.10 M * Veq mL) / (25 mL + Veq mL)
Using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
pH = pKa + log ((0.10 M * Veq mL) / (25 mL + Veq mL)) / ((0.10 M * 25 mL) / (25 mL + Veq mL))
With pKa = 4.76 as before, substitute the values into the equation to calculate the pH at the equivalence point.
(c) pH at the midpoint of the titration:
At the midpoint of the titration, an equal volume (Veq / 2 mL) of NaOH has been added to the solution, which means half of the HAc has been neutralized.
[HAc] = (0.10 M * 25 mL) / (25 mL + Veq / 2 mL)
[A-] = (0.10 M * Veq / 2 mL) / (25 mL + Veq / 2 mL)
Similarly, use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
pH = pKa + log ((0.10 M * Veq / 2 mL) / (25 mL + Veq / 2 mL)) / ((0.10 M * 25 mL) / (25 mL + Veq / 2 mL))
Substitute the values into the equation to calculate the pH at the midpoint of the titration.
Please note that these calculations are based on the assumption of complete ionization and neglecting the volume change due to the addition of NaOH.
To determine the pH at different points during the titration of HAc with NaOH, we need to consider the acid-base reactions that occur and take into account the concentrations of both the acid and the base.
(a) pH at the beginning of titration:
At the beginning of the titration, before any NaOH is added, we have only the HAc present. HAc is a weak acid, and it undergoes partial ionization in water according to the equation:
HAc (aq) ⇌ H+ (aq) + Ac- (aq)
Using the equation for the acidity constant Ka = [H+][Ac-]/[HAc], where [ ] represents concentration, we can express the relationship between the concentrations of HAc, H+, and Ac-. In this case, [H+] = [Ac-], and [HAc] is given as 0.10 M.
Assuming that the dissociation of HAc is negligible compared to its initial concentration, we can approximate the concentration of [H+] as 0.10 M. Taking the -log of [H+], we find that the pH at the beginning of titration is approximately 1.0.
(b) pH at the equivalence point of the titration:
The equivalence point of the titration occurs when an equal number of moles of HAc and NaOH are reacted. At this point, all the HAc has been converted to its conjugate base, Ac-, as well as water being formed from the reaction of H+ and OH-.
The reaction that occurs at the equivalence point is: HAc + NaOH ⇌ NaAc + H2O
Since NaAc is a sodium salt of a weak acid, it hydrolyzes in water to produce Ac- and OH-. The concentration of OH- in the solution is determined by the hydrolysis of NaAc. The pH at the equivalence point can be calculated by considering the hydrolysis reaction of NaAc.
(c) pH at the midpoint of the titration:
The midpoint of the titration occurs when the moles of HAc reacted with NaOH are half the moles needed to reach the equivalence point. At this point, the solution contains equal amounts of HAc and its conjugate base, Ac-. This is important because Ac- is a buffer solution.
A buffer solution resists changes in pH when small amounts of acid or base are added. The pH at the midpoint of the titration depends on the pKa of the acid. Since HAc is a weak acid, we can calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log([Ac-]/[HAc])
By determining the pKa value of HAc, which is 4.74, and substituting the concentrations of Ac- and HAc at the midpoint of the titration, we can calculate the pH. Given that at the midpoint of the titration, half of the initial HAc (12.5 mL) has reacted and 12.5 mL of NaOH has been added, we can determine the concentrations of Ac- and HAc to calculate the pH.