What is the concentration of the SCN- ion in the stock solution of 2.00 x 10-3 M NaSCN in 0.10M HNO3?
(SCN^-) = (NaSCN)
To find the concentration of the SCN- ion in the stock solution, we need to consider the dissociation of NaSCN in water. When NaSCN dissociates, it forms Na+ ions and SCN- ions.
The chemical equation for the dissociation of NaSCN is:
NaSCN ⇌ Na+ + SCN-
Given that the stock solution has a concentration of 2.00 x 10-3 M NaSCN, it means that there are 2.00 x 10-3 moles of NaSCN per liter of solution.
Now, let's assume that x represents the concentration of SCN- ions, both from the dissociation of NaSCN and the presence of HNO3. We'll now set up an equation based on the ion concentrations:
[Na+] + [SCN-] = 2.00 x 10-3 M
Since Na+ comes from NaSCN, and assuming complete dissociation, the concentration of Na+ equals the concentration of NaSCN, which is 2.00 x 10-3 M.
[HNO3] is given as 0.10 M, which is the concentration of HNO3 in the solution.
Taking this into account, the equation becomes:
2.00 x 10-3 M + x = 0.10 M
Simplifying the equation:
x = 0.10 M - 2.00 x 10-3 M
x = 0.098 M
Therefore, the concentration of the SCN- ion in the stock solution is approximately 0.098 M.