Calculate the pH of a solution containing 0.0300M potassium hydrogen tartrate and 0.018M dipotassium tatrate.

Question

The ka values of tartaric acid is 9.20*10^-4 and 4.31*10^-5

Answer 1

The equation is: pH = pKa + log10([A-]/[HA])

For [HA] (HC4H4O6):

1. [HA] = 0.0300 M (the concentration of potassium hydrogen tartrate).

For [A-] (C4H4O6^2-):

Since dipotassium tartrate (K2C4H4O6) is the conjugate base of the weak acid (H2C4H4O6), we need to consider its contribution as well. The concentration of dipotassium tartrate is 0.018 M, and it will dissociate into two moles of C4H4O6^2- for every mole of K2C4H4O6.

1. [A-] = 2 * 0.018 M = 0.036 M

plug these values into the Henderson-Hasselbalch equation:

1. pH = pKa + log10([A-]/[HA])
2. pH = 3.5 + log10(0.036 M / 0.0300 M)

Calculate the ratio:

pH = 3.5 + log10(1.2)

pH = 3.5 + 0.0792

pH ≈ 3.5792