Calculate the pH at the halfway point and at the equivalence point for each of the following titrations.100.0 mL of 0.56 M HCl titrated by 0.28 M NaOH
halfway point
equivalence point
To calculate the pH at the halfway point and at the equivalence point for the given titration, we need to consider the stoichiometry of the reaction.
The balanced equation for the reaction between HCl and NaOH is:
HCl + NaOH -> NaCl + H2O
1. Halfway point: At the halfway point, equal moles of HCl and NaOH have reacted. This means that half of the initial moles of HCl have reacted. To find the moles of HCl that have reacted, we can use the following formula:
moles = concentration x volume
Initial moles of HCl = 0.56 M x 0.100 L = 0.056 mol
Since we have reached the halfway point, half of the initial HCl has reacted, and the remaining HCl is equal to:
Remaining moles of HCl = 0.056 mol / 2 = 0.028 mol
To calculate the concentration of HCl, we divide the remaining moles by the volume of NaOH added to reach the halfway point. Since the volume of NaOH added is half of the total volume (100.0 mL), the volume of NaOH added is 50.0 mL or 0.050 L.
HCl concentration at halfway point = 0.028 mol / 0.050 L = 0.56 M
Since HCl is a strong acid, it fully dissociates in water, and the concentration of H3O+ ions is equal to the concentration of HCl. Therefore, the pH at the halfway point is equal to the negative logarithm (base 10) of the HCl concentration:
pH = -log[H3O+]
pH = -log[0.56]
pH ≈ 0.25
2. Equivalence point: At the equivalence point, the moles of acid (HCl) are equal to the moles of base (NaOH). Since the balanced equation has a 1:1 stoichiometry between HCl and NaOH, this means that all of the HCl has reacted, and we have formed an equal number of moles of NaCl and H2O.
At the equivalence point, the solution contains only the salt NaCl. NaCl is a neutral salt, so it does not contribute to the pH of the solution. Therefore, the pH at the equivalence point will be neutral, around pH 7.
To calculate the pH at the halfway point and at the equivalence point for a titration, we need to consider the stoichiometry of the reaction and the concentrations of the reactants.
In this case, we have the titration of HCl with NaOH:
HCl + NaOH → NaCl + H2O
Step 1: Halfway Point
At the halfway point of the titration, equal amounts of HCl and NaOH have reacted. This means that half of the HCl has been neutralized by NaOH.
To determine the volume of NaOH needed to reach the halfway point, we can use the equation:
(V1)(C1) = (V2)(C2)
(V1) (0.56 M) = (V2) (0.28 M)
V1 = (0.28 M)(V2) / (0.56 M)
Since the initial volume of HCl is 100.0 mL, and we're looking for the halfway point, half of it would be 50.0 mL. We can substitute this value into the equation to solve for V2:
(0.28 M)(V2) / (0.56 M) = 50.0 mL
V2 ≈ 25.0 mL
So, at the halfway point, 25.0 mL of 0.28 M NaOH has been added to 100.0 mL of 0.56 M HCl.
Now, we can calculate the moles of HCl and NaOH at the halfway point:
moles of HCl = (0.56 M)(0.100 L) = 0.056 mol
moles of NaOH = (0.28 M)(0.025 L) = 0.007 mol
Since NaOH is the limiting reactant (less moles), it will be completely consumed, leaving an excess of HCl. Therefore, the pH at the halfway point will be determined by the excess HCl.
To calculate the pH, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
In this case, HCl is a strong acid, so it completely dissociates in water, and we have a high excess of HCl. As a result, [A-] (concentration of conjugate base) is negligible compared to [HA] (concentration of acid). Therefore, the pH at the halfway point would essentially be determined by the concentration of HCl.
pH = -log10 [HCl]
To find the concentration of HCl at the halfway point, we need to account for the dilution caused by the addition of NaOH. Since the total volume after adding NaOH is 125.0 mL (100.0 mL HCl + 25.0 mL NaOH), we can calculate the concentration of HCl at this volume:
[HCl] = (0.056 mol) / (0.125 L)
Now, we can calculate the pH at the halfway point:
pH = -log10 [HCl] = -log10 [(0.056 mol) / (0.125 L)]
Step 2: Equivalence Point
At the equivalence point of the titration, the moles of HCl and NaOH are in a 1:1 ratio, meaning the acid has been completely neutralized by the base. In this case, the equivalence point occurs when 0.056 mol of NaOH has been added.
Since NaOH is a strong base, it completely dissociates in water, resulting in the production of OH- ions. Therefore, at the equivalence point, the solution will be basic, and the pH can be calculated using the concentration of hydroxide ions [OH-].
To find the concentration of OH- at the equivalence point, we can calculate the moles of OH-:
moles of OH- = (0.28 M)(0.025 L)
Now, we can calculate the concentration of OH- at the equivalence point:
[OH-] = (moles of OH-) / (total volume at equivalence point)
The total volume at the equivalence point can be obtained by adding the initial volume of HCl (100.0 mL) to the volume of NaOH required to reach the equivalence point (V2 ≈ 50.0 mL):
total volume at equivalence point = 100.0 mL + 50.0 mL = 150.0 mL
Finally, we can calculate the pH at the equivalence point:
pOH = -log10 [OH-]
pH = 14 - pOH
Now, substitute the calculated [OH-] into these equations to determine the pH at the equivalence point.