What is the fraction of NTA present after HT2- has been brought to equilibrium with solid PbCO3 at pH 7.00, in a medium in which [HCO3-] = 1.25E-3 M?
To determine the fraction of NTA present after HT2- has been brought to equilibrium with solid PbCO3 at pH 7.00, we need to set up the chemical equation and use the concept of equilibrium and dissociation.
First, let's write the chemical equation:
HT2- + PbCO3 ⇌ H+ + T2- + PbCO3
From the given information, we know that the medium has a pH of 7.00, which implies that [H+] = 10^-7 M. We are also given the concentration of HCO3- to be 1.25E-3 M.
To find the fraction of NTA present, we need to calculate its concentration in terms of [T2-]. To do this, we need to consider the equilibrium expression and the fact that HT2- dissociates as follows:
HT2- ⇌ H+ + T2-
Let's assign x as the concentration of T2- (and H+) in this equilibrium. Since the stoichiometric coefficient of T2- is 1 in the equilibrium equation, [T2-] = x, and [H+] will also be equal to x (since we have a 1:1 ratio due to dissociation).
Now, let's consider the equilibrium expression for HT2-:
K = [H+][T2-] / [HT2-]
Since [H+] = [T2-] = x, we can rewrite the expression as:
K = x^2 / [HT2-] ---- (1)
The value of K is the equilibrium constant, which depends on the particular reaction. Unfortunately, the given information does not provide the value of K.
However, we can make an assumption that [HT2-] is much larger than x. This assumption is valid if the initial concentration of HT2- is significantly larger than the concentration of T2- at equilibrium.
With this assumption, we can neglect x in comparison to [HT2-] when x is calculated and solve for [HT2-].
Let's assume that [HT2-] = C. In the reaction, HT2- reacts with PbCO3 to form H+ and T2-. Therefore, C moles of HT2- will react with C moles of PbCO3 to produce C moles of T2-. This implies that C moles of T2- will be present in the solution.
So, the fraction of NTA present after equilibrium is
[T2-] / [HT2-] = C / C = 1.
This means that 100% of the HT2- has reacted to form T2- at equilibrium.
Note: This explanation assumes that the equilibrium constant (K) remains constant throughout the reaction. The value of K can vary depending on factors such as temperature and pressure.