Some of 0.2M of acetic acid are mixed wit 50ml of 0.2 sodium acetate solution, what will be the pH of the mixture?? (Ka=1.85 ×10*-10 for acetic acid).

To determine the pH of the mixture, we need to calculate the concentration of acetic acid and acetate ions in the solution after mixing.

Step 1: Calculate the moles of acetic acid in 0.2 M solution.
The volume of the 0.2 M acetic acid solution is not provided, so let's assume it is also 50 mL. Therefore, the number of moles of acetic acid is:
Moles of acetic acid = concentration * volume
= 0.2 mol/L * 0.05 L
= 0.01 mol

Step 2: Calculate the concentration of acetic acid and acetate ions after mixing.
Since the acetic acid and sodium acetate react to form a buffer solution, we need to consider the reaction between acetic acid and sodium acetate:
CH3COOH + NaCH3COO ⇌ CH3COONa + H2O

From the balanced equation, we can deduce that every 1 mole of acetic acid reacts with 1 mole of sodium acetate to form 1 mole of acetate ions (CH3COO-). Therefore, the number of moles of acetate ions in the mixture is also 0.01 mol.

Step 3: Calculate the total volume of the mixture.
To determine the concentration of the resulting solution, we need to know the total volume after mixing. However, the question does not provide this information. Let's assume the volume remains 50 mL.

Step 4: Calculate the concentration of acetic acid and acetate ions in the mixture.
The final concentration of acetic acid is given by:
Concentration of acetic acid = Moles of acetic acid / Total volume
= 0.01 mol / 0.05 L
= 0.2 M

Similarly, the final concentration of acetate ions is:
Concentration of acetate ions = Moles of acetate ions / Total volume
= 0.01 mol / 0.05 L
= 0.2 M

Step 5: Calculate the ratio of acetate ions to acetic acid.
Since we have equal concentrations of acetic acid and acetate ions (0.2 M each), the ratio of acetate ions to acetic acid is 1:1.

Step 6: Calculate the pH using the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log(base 10) (concentration of acetate ions / concentration of acetic acid)

Considering that the Ka value for acetic acid is 1.85 × 10^-5 (pKa = -log10 (Ka) = -log10 (1.85 × 10^-5) = 4.73), we can substitute the values into the equation:

pH = 4.73 + log(base 10) (0.2 M / 0.2 M)
= 4.73

Therefore, the pH of the mixture is 4.73.