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
13 years ago
11 months ago
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.