Verified answer

1)

The Henderson-Hasselbalch equation is generally used for buffer solutions--solutions that contain an acid and its conjugate base. Therefore, the [HA] and [A-] are generally given in the problem (for example, if you have 0.383 M CH3COOH and 0.225 M CH3COONa...). HA refers to the weak acid while A- refers to the conjugate base (acetic acid would be HA and sodium acetate would be A-, for example).

2)

From the equation:

pH = pKa + log[A-/HA]

You can see that pH is affected by the concentration of A- and HA. When [HA] and [A-] are equal, the ratio between the two is 1. Therefore, log(1) = 0, so the pH = pKa.

Whenever log(x) < 1, the number is negative. Whenever log(x) > 1, the number is positive.

You can see that if [A-] is greater than [HA], the A-/HA > 1. When you add the number to the pKa, the pH is increased, and therefore, the solution more basic than its pKa. Keep in mind that this does not mean that the overall solution is basic, it just means that the pH > pKa, implying more conjugate base than acid. If [HA] is greater than [A-], then the A-/HA < 1, and therefore you would subtract that number from the pKa, and thus giving you a lower pH (pH < pKa). This just means that there is more acid than conjugate base.

For example, if you have CH3COOH and CH3COONa (pKa = 4.74). If you have 0.500 M CH3COONa and 0.500 M CH3COOH, you can see that:

pH = 4.74 + log(0.5/0.5) = 4.74 + 0 = 4.74

If you have 0.750 M CH3COONa and 0.250 M CH3COOH:

pH = 4.74 + log(0.750/0.250) = 4.74 + 0.48 = 5.22

If you have 0.250 M CH3COONa and 0.750 M CH3COOH:

pH = 4.74 + log(0.250/0.750) = 4.74 - 0.48 = 4.26