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
NaOH + HCl ==> NaCl + H2O
These are 1:1 reactions.
Calculate volume for equivalence point, then take half that for the half way point.
Make an ICE chart for the half way point, and solve for H^+, then convert to pH.
For the pH at the equivalence point, the pH will be determined by the hydrolysis of the salt. What is the salt? How is it hydrolyzed?
To calculate the pH at the halfway point and the equivalence point of the given titration, we need to consider the stoichiometry and the reaction that occurs during the titration.
In this case, HCl is a strong acid, and NaOH is a strong base. The reaction between them can be represented as follows:
HCl + NaOH → NaCl + H2O
At the halfway point, equal moles of HCl and NaOH will react, which means that half of the HCl has been neutralized by the NaOH. Therefore, the number of moles of HCl remaining will be half of the initial moles of HCl.
To find the halfway point volume, we can use the formula:
Volume of base at halfway point = Volume of acid × (Molar concentration of acid / Molar concentration of base)
Given that the initial volume of HCl is 100.0 mL (0.100 L) and the molar concentration of HCl is 0.56 M, and the molar concentration of NaOH is 0.28 M, we can calculate the halfway point volume:
Volume of base at halfway point = 0.100 L × (0.56 M / 0.28 M)
= 0.200 L
Now, we have the volume of NaOH required to reach the halfway point, which is 0.200 L. Since the molar ratio between HCl and NaOH is 1:1, this means that half of the initial moles of HCl has reacted. Therefore, the remaining moles of HCl will also be half:
Moles of HCl remaining = (Initial moles of HCl) / 2
= (0.100 L × 0.56 M) / 2
= 0.028 moles
To calculate the pH at the halfway point, we need to determine the concentration of the remaining HCl. We can use the equation:
Molarity (M) = Moles / Volume (L)
Therefore,
Molarity of HCl at halfway point = 0.028 Moles / 0.100 L
= 0.28 M
Since HCl is a strong acid, it fully ionizes in water, resulting in H+ ions. Therefore, the pH of a 0.28 M HCl solution is equal to the negative logarithm base 10 of its concentration:
pH at the halfway point = -log10(0.28)
≈ 0.55
Now, let's calculate the pH at the equivalence point, where stoichiometrically, all of the HCl has reacted with NaOH. At the equivalence point, we have added an equal amount of NaOH to exactly neutralize the HCl.
The volume of NaOH required to reach the equivalence point is calculated using the same formula as before. However, at the equivalence point, the volume of base added will be equal to the volume of acid initially added:
Volume of base at equivalence point = 0.100 L × (0.56 M / 0.28 M)
= 0.200 L
Since all the HCl has been neutralized, the moles of remaining HCl will be zero. Therefore, the concentration of HCl at the equivalence point is zero. In this situation, the solution will be neutral, resulting in a pH of 7.
So, at the halfway point, the pH is approximately 0.55, and at the equivalence point, the pH is 7.