The expression you provided is a quadratic equation involving the bicarbonate ion $[HCO_3^-]$, carbon dioxide $[CO_2^*]$, dissociation constants $K_1$, $K_2$, and the dissolved inorganic carbon (DIC).

The expression can be rearranged as follows:

[ [HCO_3^-]^2 + \left( \frac{K_1 [CO_2^]}{K_2} \right)[HCO_3^-] - \left( \frac{K_1 [CO_2^]}{K_2} \right)(DIC - [CO_2^*]) = 0 ]

Given the constants:

• $K_1 = 1.25 \times 10^{-6}$
• $K_2 = 8.55 \times 10^{-10}$
• DIC = $2.03 \times 10^{-3}$
• $[CO_2^*] = 13.14 \times 10^{-6}$

This is a quadratic equation of the form:

$Ax^2 + Bx + C = 0$

where:

• $A = 1$
• $B = \left(\frac{K_1 [CO_2^*]}{K_2}\right)$
• $C = -\left(\frac{K_1 [CO_2^*]}{K_2}\right)(DIC - [CO_2^*])$

Let me calculate the values of $B$ and $C$, and then solve the quadratic equation.The quadratic equation has the following coefficients:

• $B = 0.01921$
• $C = -3.874 \times 10^{-5}$

The solutions (roots) for $[HCO_3^-]$ are:

• $[HCO_3^-]_1 = 0.00184 \, \text{mol/L}$
• $[HCO_3^-]_2 = -0.02105 \, \text{mol/L}$ (which is not physically meaningful because concentrations cannot be negative).

Therefore, the valid concentration of bicarbonate ion $[HCO_3^-]$ is approximately:

$[HCO_3^-] = 0.00184 \, \text{mol/L}$

Would you like further details or have any questions?

Here are five related questions to expand on this:

1. How are $K_1$ and $K_2$ values related to acid-base equilibria?
2. What is the role of DIC in oceanic carbon cycles?
3. How does the concentration of bicarbonate ions affect pH levels?
4. What would happen if the concentration of $[CO_2^*]$ changes significantly?
5. How can this type of calculation be used in environmental chemistry?

Tip: Always check if the discriminant in a quadratic equation is positive to ensure real and meaningful solutions.