Question

Citric acid, also known as vitamin C, is a triprotic acid with the formula H3C6H5O7. The pKa values are 3.13, 4.76, and 6.40. What will be the approximate pH of a solution of sodium monohydrogencitrate, Na2HC6H5O7?

Answer 1

Answer:

pH = 5.58

Explanation:

Given that :

Citric acid  is a triprotic acid with the formula $H_3C_6H_5O_7$ . Tripotic acids are acids that have the tendency to yield three protons per molecule during the phases of dissociation, therefore they have three pka.

For Citric acid; the pka values are:

$pKa _1 = 3.13\\pKa_2 = 4.76\\pKa_3 = 6.40$

So, the pH of sodiummonohydrogencitrate with the chemical formula $Na_2HC_6H_5O_7$ which can act as an acid and a base (i.e it is said to be amphoteric) can be calculated as :

$pH = \frac{pKa_2+pKa_3}{2}$

$pH = \frac{4.76+6.40}{2}$

pH = 5.58

Therefore, the approximate pH of a solution of sodium monohydrogencitrate = 5.58