determine the pH of a solution made by mixing 8 ml of 1.10M acetic acid and 2ml 0.9 M sodium acetate solution. pKa for acetic acid CH3COOH is 4.745.

Can someone please explain

Acetic acid is a weak acid; sodium acetate is a salt of a weak acid. The combination makes a buffered solution. You calculate pH from the Henderson-Hasselbalch equation.

Sure! To determine the pH of the solution, we need to use the Henderson-Hasselbalch equation, which relates the pH of a solution to the concentration of the acid and its conjugate base.

The Henderson-Hasselbalch equation is:

pH = pKa + log([conjugate base]/[acid])

In this case, acetic acid (CH3COOH) is the acid and sodium acetate (CH3COONa) is its conjugate base.

Step 1: Calculate the moles of acetic acid and sodium acetate.

Moles of acetic acid = volume (L) x concentration (M)
= 0.008 L x 1.10 M
= 0.0088 moles

Moles of sodium acetate = volume (L) x concentration (M)
= 0.002 L x 0.9 M
= 0.0018 moles

Step 2: Calculate the concentrations of acetic acid and sodium acetate.

Concentration of acetic acid = moles/volume (L)
= 0.0088 moles / 0.01 L
= 0.88 M

Concentration of sodium acetate = moles/volume (L)
= 0.0018 moles / 0.01 L
= 0.18 M

Step 3: Substitute the values into the Henderson-Hasselbalch equation.

pH = pKa + log([0.18]/[0.88])

Step 4: Solve for pH.

pH = 4.745 + log(0.18/0.88)
= 4.745 - 0.66
= 4.085

Therefore, the pH of the solution made by mixing 8 ml of 1.10 M acetic acid and 2 ml of 0.9 M sodium acetate solution is approximately 4.085.