Question

What is ph of .530 M solution of carbonic acid that has acid dissociation of 4.4 x 10^-7

Answer 1

pH = 3.32

Step-by-step explanation

Given

Acid dissociation of carbonic acid (Ka) = 4.4x10^-7

Concentration = 0.530 M

Solution

The carbonic acid dissociates into:

`\begin{gathered} H_2CO_3\rightarrow\text{ H}^{+\text{ }}+\text{ HCO3}^- \\ \end{gathered}`
`\begin{gathered} Ka\text{ = }([H^+][A^-])/([HA]) \\ [H^+]\text{ = \lbrack A}^-] \\ Ka\text{ x \lbrack HA\rbrack= \lbrack H}^+]^2 \\ 4.4\text{ x 10}^(-7)\text{ x 0.530 = \lbrack H}^+]^2 \\ 2.332\text{ x 10}^(-7)\text{ = \lbrack H}^+]^2 \\ 4.816\text{ x 10}^(-4)\text{ = \lbrack H}^+] \\ \\ \end{gathered}`

Now we can calculate the pH

pH = -log[H+]

pH = -log[4.816x10^-4]

pH = 3.32