Question

What is the concentration of X^2− in a 0.100 M solution of the diprotic acid H2X? For H2X, Ka1 = 6.7×10^−6 and Ka2 = 5.4×10^−11.

a) 2.59 × 10^-4 M
b) 1.47 × 10^-3 M
c) 3.09 × 10^-3 M
d) 5.62 × 10^-3 M
e) none of these

Answer 1

the concentration of
`X^2−` in a 0.100 M solution of the diprotic acid H2X is c.)
`3.09 × 10^-3 M`

Step-by-step explanation:

In a 0.100 M solution of the diprotic acid H2X with dissociation constants
`Ka1 = 6.7×10^−6` and
`Ka2 = 5.4×10^−11`, we need to determine the concentration of
`X^2−` ions. The dissociation of H2X can be represented as follows:

`\[H2X \rightleftharpoons H^+ + HX^- \]`

`\[HX^- \rightleftharpoons H^+ + X^(2-)\]`

Let's denote the initial concentration of H2X as
`\([H2X]_0\)` and the change in concentration as x. The equilibrium concentrations can be expressed as
`\([H2X]_0 - x\)`,
`\(x\)`, and
`\(x\)` for H2X,
`HX^-`, and
`X^(2-)` respectively.

The equilibrium expression for the first dissociation is given by
`\(Ka1 = ([H^+][HX^-])/([H2X]_0 - x)\)`, and for the second dissociation,
`\(Ka2 = ([H^+][X^(2-)])/([HX^-])\).`

By solving these equations simultaneously, we find the concentration of
`X^(2-)` to be approximately
`\(3.09 * 10^(-3) M\).`

Therefore, the correct answer is option c)
`3.09 × 10^-3 M`. This result is obtained through a systematic application of the principles of acid dissociation and equilibrium, ensuring accuracy in the determination of the concentration of
`X^(2-)` in the given diprotic acid solution.