To determine the concentration of the riciculite ion in a 2.376 M solution of riciculous acid, we need to consider the dissociation of the acid.
Riciculous acid (H2RIC) dissociates in two steps, forming two riciculite ions (RIC-) successively:
Step 1: H2RIC ⇌ H+ + HRIC-
Step 2: HRIC- ⇌ H+ + RIC-
Given:
Ka1 = 2.01 × 10^(-5)
Ka2 = 1.96 × 10^(-10)
Initial concentration of riciculous acid (H2RIC) = 2.376 M
Let's assign 'x' as the concentration of H+ ions and HRIC- ions formed in the first dissociation step.
Using the Ka1 expression, we can write the equilibrium expression for the first dissociation:
Ka1 = [H+][HRIC-] / [H2RIC]
Since the concentration of H2RIC is much larger than 'x' (the concentration of H+ and HRIC- ions formed), we can approximate [H2RIC] ≈ 2.376 M.
Substituting the given values, we have:
2.01 × 10^(-5) = x^2 / 2.376
Solving for 'x', we find:
x^2 = 2.01 × 10^(-5) * 2.376
x^2 = 4.7736 × 10^(-5)
x ≈ 0.006906 M
Now, for the second dissociation step, we can assume that 'x' is the concentration of H+ ions and RIC- ions formed:
[H+][RIC-] / [HRIC-]
Using the Ka2 expression, we have:
Ka2 = [H+][RIC-] / [HRIC-]
1.96 × 10^(-10) = x^2 / (2.376 - x)
Since the concentration of HRIC- is much larger than 'x', we can approximate [HRIC-] ≈ 2.376 M.
Substituting the given values, we get:
1.96 × 10^(-10) = x^2 / (2.376 - x)
Solving for 'x', we find:
x^2 = 1.96 × 10^(-10) * (2.376 - x)
x^2 = 3.8496 × 10^(-10) - 1.96 × 10^(-10)x
x^2 + 1.96 × 10^(-10)x - 3.8496 × 10^(-10) = 0
Using the quadratic formula, we can solve for 'x':
x ≈ 0.000586 M (approximately)
Therefore, the concentration of the riciculite ion (RIC-) in the 2.376 M solution of riciculous acid is approximately 0.000586 M.