(a) Calculate the pH of a 0.15M acetate buffer (a solution containing 0.1 mol/L acetic acid and 0.1 mol/L sodium acetate), given that the pKa of acetic acid is 4.7.
(b) What would be the pH value after adding 0.05 mmol of sodium to 1l of 0.2 mol/L acetate buffer?
(a) To calculate the pH of an acetate buffer solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Where pH is the desired pH value, pKa is the dissociation constant of the weak acid (acetic acid in this case), [A-] is the concentration of the conjugate base (sodium acetate), and [HA] is the concentration of the weak acid (acetic acid).
In this case, the concentration of acetic acid (weak acid) and sodium acetate (conjugate base) is given as 0.1 mol/L each. The pKa of acetic acid is 4.7.
Substituting these values into the Henderson-Hasselbalch equation:
pH = 4.7 + log ([0.1 mol/L] / [0.1 mol/L])
Simplifying:
pH = 4.7 + log (1)
The log of 1 is 0, so:
pH = 4.7
Therefore, the pH of the 0.15M acetate buffer is 4.7.
(b) To determine the new pH after adding 0.05 mmol of sodium to 1L of a 0.2 mol/L acetate buffer, we need to consider the effect of dilution on the final pH.
First, calculate the moles of sodium added:
0.05 mmol = 0.05 x 10^-3 mol
Next, calculate the new concentration of sodium acetate after adding the sodium:
Initial concentration of sodium acetate = 0.2 mol/L
Initial volume of the solution = 1L
Moles of sodium acetate initially = concentration x volume
Moles of sodium acetate initially = 0.2 mol/L x 1 L
Moles of sodium acetate initially = 0.2 mol
Final concentration of sodium acetate = (initial moles + added moles) / initial volume
Final concentration of sodium acetate = (0.2 mol + 0.05 x 10^-3 mol) / 1L
Final concentration of sodium acetate = 0.2 mol / 1L = 0.2 mol/L
Now, we can use the Henderson-Hasselbalch equation to calculate the new pH:
pH = pKa + log ([A-]/[HA])
In this case, the pKa and initial concentrations of acetic acid and sodium acetate remain the same as in part (a). However, the concentration of sodium acetate (conjugate base) has changed to 0.2 mol/L.
So, substituting these values into the Henderson-Hasselbalch equation:
pH = 4.7 + log ([0.2 mol/L] / [0.1 mol/L])
Simplifying:
pH = 4.7 + log (2)
Using the logarithm properties, we find:
pH = 4.7 + 0.3010
pH = 4.7 + 0.3010
pH = 5.001
Therefore, the pH value after adding 0.05 mmol of sodium to 1L of 0.2 mol/L acetate buffer is approximately 5.001.