Assuming that Ka is 1.85 *10-5 for acetic acid, calculate the pH at one-half the equivalence point and at the equivalence point for titration of 50mL of 0.100M acetic acid with 0.100M NaOH.

To calculate the pH at one-half the equivalence point and at the equivalence point for the titration of acetic acid with NaOH, we can use the following steps:

Step 1: Write the balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):

CH3COOH + NaOH → CH3COONa + H2O

Step 2: Determine the number of moles of acetic acid (CH3COOH) in the 50 mL solution:

Number of moles = concentration (M) × volume (L)

Number of moles = 0.100 M × 0.050 L

Step 3: Determine the volume of NaOH needed to reach the equivalence point. Since acetic acid and sodium hydroxide react in a 1:1 ratio, the volume of NaOH needed will be equal to the volume of acetic acid solution:

Volume NaOH = 50 mL

Step 4: At one-half the equivalence point, half of the acetic acid has reacted with sodium hydroxide. This means that half of the acetic acid (CH3COOH) has been converted to sodium acetate (CH3COONa).

Since the reaction is 1:1, the concentration of acetic acid remaining at one-half the equivalence point is:

Concentration (CH3COOH) = Initial concentration (CH3COOH) - Concentration (CH3COONa)

Concentration (CH3COOH) = 0.100 M - (0.100 M / 2)

Step 5: Calculate the pH at one-half the equivalence point using the Henderson-Hasselbalch equation:

pH = pKa + log10([A-]/[HA])

Where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base (CH3COO-) and [HA] is the concentration of the weak acid (CH3COOH).

Using the given Ka value of 1.85 × 10^-5, we can calculate the pKa:

pKa = -log10(Ka)
pKa = -log10(1.85 × 10^-5)

Using logarithmic properties:

pKa = -log10(1.85) - log10(10^-5)
pKa = -log10(1.85) + 5

Now we can substitute the known values into the Henderson-Hasselbalch equation:

pH = pKa + log10([A-]/[HA])
pH = (pKa + log10([A-]) - log10([HA])
pH = (pKa + log10([A-])) - log10([HA])

At one-half the equivalence point, the concentration of sodium acetate (CH3COO-) is equal to the concentration of acetic acid (CH3COOH), and therefore:

[A-] = [HA]

pH = (pKa + log10([A-])) - log10([HA])
pH = (pKa + log10([HA])) - log10([HA])
pH = pKa

Hence, at one-half the equivalence point, the pH of the solution is equal to the pKa value calculated in Step 5.

Step 6: At the equivalence point, all of the acetic acid (CH3COOH) has reacted with the sodium hydroxide (NaOH) to form sodium acetate (CH3COONa). This means that no acetic acid is remaining in the solution.

Therefore, the concentration of acetic acid (CH3COOH) at the equivalence point is:

Concentration (CH3COOH) = 0 M

In this case, since there is no acetic acid remaining, the pH at the equivalence point is determined by the hydrolysis of sodium acetate (CH3COONa).

Step 7: The pH of a sodium acetate solution can be calculated using the equation:

pH = pKa + log10([A-]/[HA])

Since there is no acetic acid remaining at the equivalence point, the concentration of [HA] is 0, and the equation simplifies to:

pH = pKa + log10([A-]/0)

Since log10(0) is undefined, the pH at the equivalence point is indeterminate.

To summarize:
- At one-half the equivalence point, the pH is equal to the pKa value.
- At the equivalence point, the pH is indeterminate.

To calculate the pH at one-half the equivalence point and at the equivalence point for the titration of acetic acid with NaOH, we need to determine the concentration of acetic acid and NaOH at those points.

First, let's calculate the concentration of acetic acid (CH3COOH) in moles:

moles of acetic acid = volume (L) × concentration (M)
= 0.050 L × 0.100 M
= 0.005 moles

Now, at half the equivalence point, the number of moles of acetic acid that have reacted with NaOH is equal to half of the initial moles of acetic acid:

moles of acetic acid reacted = (1/2) × 0.005 moles
= 0.0025 moles

Since NaOH reacts with acetic acid in a 1:1 ratio, the number of moles of NaOH required to react with 0.0025 moles of acetic acid is also 0.0025 moles.

At the half-equivalence point, the concentration of acetic acid remaining is given by:

concentration of acetic acid remaining = remaining moles / volume (L)
= (0.005 - 0.0025) moles / 0.050 L
= 0.0025 moles / 0.050 L
= 0.050 M

To calculate the pH at this point, we need to consider the dissociation of acetic acid. Acetic acid (CH3COOH) dissociates into acetate ions (CH3COO-) and hydronium ions (H3O+).

The dissociation reaction of acetic acid can be represented as:
CH3COOH + H2O ⇌ CH3COO- + H3O+

The equilibrium constant for this reaction is given by Ka, which is 1.85 × 10^-5.

Using the Ka expression, we can write the equation for the dissociation of acetic acid:

Ka = [CH3COO-] × [H3O+] / [CH3COOH]

Since the concentration of CH3COO- and H3O+ at half the equivalence point is the same, and [CH3COOH] is equal to the concentration of acetic acid remaining, we can rewrite the equation as:

Ka = [H3O+]^2 / [CH3COOH]

Rearranging the equation, we get:

[H3O+]^2 = Ka × [CH3COOH]

Taking the square root of both sides, we have:

[H3O+] = sqrt(Ka × [CH3COOH])

Substituting the values, we get:

[H3O+] = sqrt(1.85 × 10^-5 × 0.050)

Now, calculate the pH using the formula:

pH = -log[H3O+]

Substituting the value for [H3O+], we can determine the pH at one-half the equivalence point.

Similarly, at the equivalence point, all the acetic acid has reacted with NaOH. Therefore, the concentration of acetic acid remaining is zero. In this case, the concentration of acetate ions (CH3COO-) is equal to the concentration of NaOH used.

Since NaOH and CH3COOH react in a 1:1 ratio, the concentration of CH3COO- is:

concentration of CH3COO- = (0.100 M) × (volume of NaOH used / total volume)

Since 50 mL of 0.100 M NaOH is used, the total volume at the equivalence point is 100 mL. Substituting these values, we can determine the concentration of CH3COO-.

To calculate the pH at the equivalence point, we can use the formula:

pH = -log[OH-]

Since NaOH is a strong base, it fully dissociates into Na+ and OH-. Therefore, the concentration of OH- is equal to the concentration of NaOH used. We can substitute this value into the pH equation to find the pH at the equivalence point.

By following these steps and performing the calculations, we can determine the pH at one-half the equivalence point and at the equivalence point for the given titration.

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