what will be the pH of a solution prepared by mixing 100 mL of 05M acetic acid with 50 mL of 05M NaOH The Ka of acetic acid is 1810^-5

To find the pH of the solution, we need to calculate the concentration of acetic acid and acetate ion after the neutralization reaction between acetic acid and NaOH.

First, let's calculate the number of moles of acetic acid and NaOH:

Number of moles of acetic acid = volume (in L) × concentration = 0.100 L × 0.5 M = 0.050 moles

Number of moles of NaOH = volume (in L) × concentration = 0.050 L × 0.5 M = 0.025 moles

Since acetic acid and NaOH react in a 1:1 ratio, the number of moles of acetic acid left after the neutralization reaction is given by:

Number of moles of acetic acid remaining = number of moles of acetic acid - number of moles of NaOH = 0.050 moles - 0.025 moles = 0.025 moles

The concentration of acetic acid remaining can be calculated as follows:

Concentration of acetic acid remaining = number of moles of acetic acid remaining / volume (in L) = 0.025 moles / (0.100 L + 0.050 L) = 0.025 moles / 0.150 L = 0.167 M

Now, we can set up an ICE table to calculate the concentration of acetate ion (CH3COO-) in the solution:

Initial concentration (M) | Change in concentration (M) | Equilibrium concentration (M) Acetic acid (CH3COOH) | 0.167 | -x | 0.167 - x NaOH | 0.050 | -x | 0.050 - x Acetate ion (CH3COO-) | 0 | +x | x

Since acetic acid is a weak acid, it partially dissociates in water, and the equilibrium expression for the dissociation reaction is:

CH3COOH + H2O ⇌ CH3COO- + H3O+

The Ka expression for acetic acid is:

Ka = [CH3COO-][H3O+] / [CH3COOH]

Since acetic acid and NaOH react in a 1:1 ratio, the concentration of acetate ion is equal to the concentration of hydroxide ions (OH-) formed in the reaction.

Therefore, we can rewrite the Ka expression as:

Ka = [OH-][H3O+] / [CH3COOH]

Substituting the known values:

1.8 × 10^-5 = x × x / 0.167 - x

Since x is much smaller than 0.167, we can approximate 0.167 - x as 0.167:

1.8 × 10^-5 = x × x / 0.167

Rearranging the equation to solve for x:

x × x = 1.8 × 10^-5 × 0.167 x² = 2.994 × 10^-6 x = √(2.994 × 10^-6) x = 0.001732

The concentration of acetate ion (CH3COO-) in the solution is 0.001732 M.

Since acetic acid and acetate ion form a conjugate acid-base pair, we can use the equation:

pH = pKa + log([CH3COO-] / [CH3COOH])

Substituting the known values:

pH = -log(1.8 × 10^-5) + log(0.001732 / 0.167) pH = 4.74 + log(0.01037) pH = 4.74 + (-1.986) pH = 2.754

Therefore, the pH of the solution prepared by mixing 100 mL of 0.5M acetic acid with 50 mL of 0.5M NaOH is approximately 2.754