Question

For the diprotic weak acid $H_2A$, $K_{a1} = 2.8 \times 10^{-6}$ and $K_{a2} = 7.5 \times 10^{-9}$. What is the pH of a 0.0600 M solution of $H_2A$? What are the equilibrium concentrations of $H_2A$ and $A^{2-}$ in this solution? $[H_2A] =$ $[A^{2-}] =$

Name: for the diprotic weak acid h2a ka1 28 times 10 6 and ka2 75 times 10 9 what is the ph of a 00600 m solution of h2a what are the equilibrium concentrations of h2a and a2 in this solution 76997
Uploaded: 2021-12-31T18:08:11-08:00
Duration: 2 min 35 s
Channel: David Collins
Description: for the diprotic weak acid h2a ka1 28 times 10 6 and ka2 75 times 10 9 what is the ph of a 00600 m solution of h2a what are the equilibrium concentrations of h2a and a2 in this solution 76997

          For the diprotic weak acid $H_2A$, $K_{a1} = 2.8 \times 10^{-6}$ and $K_{a2} = 7.5 \times 10^{-9}$.
What is the pH of a 0.0600 M solution of $H_2A$?
What are the equilibrium concentrations of $H_2A$ and $A^{2-}$ in this solution?
$[H_2A] =$
$[A^{2-}] =$

for the diprotic weak acid h2a ka1 28 times 10 6 and ka2 75 times 10 9 what is the ph of a 00600 m solution of h2a what are the equilibrium concentrations of h2a and a2 in this solution 76997

Added by Nicole H.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert David Collins

12/31/2021

Step 1

Since $H_2A$ is a diprotic acid, it will dissociate in two steps. The first dissociation is: $H_2A \rightleftharpoons H^+ + HA^-$ $K_{a1} = \frac{[H^+][HA^-]}{[H_2A]} = 2.8 \times 10^{-6}$ Let x be the concentration of $H^+$ and $HA^-$ at equilibrium. Then, the Show more…

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For the diprotic weak acid $H_2A$, $K_{a1} = 2.8 \times 10^{-6}$ and $K_{a2} = 7.5 \times 10^{-9}$. What is the pH of a 0.0600 M solution of $H_2A$? What are the equilibrium concentrations of $H_2A$ and $A^{2-}$ in this solution? $[H_2A] =$ $[A^{2-}] =$