Calculate the Concentration of Ni2+ in a soltion which prepared as mixing 50.0ml 0.0300 M Ni2+ and 50.0mlml 0.0500 M EDTA at pH= 3
At pH=3 , α4= 2.5x10-11 KNiY=4.2x1018
To calculate the concentration of Ni2+ in the solution, we will use the concept of complexation equilibrium between Ni2+ and EDTA.
The reaction between Ni2+ and EDTA can be represented as follows:
Ni2+ + Y4- ⇌ NiY2-
Given:
Volume of Ni2+ solution (V1) = 50.0 mL
Concentration of Ni2+ (C1) = 0.0300 M
Volume of EDTA solution (V2) = 50.0 mL
Concentration of EDTA (C2) = 0.0500 M
pH = 3
First, we need to convert the volumes of the solutions to liters:
V1 = 50.0 mL = 0.0500 L
V2 = 50.0 mL = 0.0500 L
Next, we need to calculate the moles of Ni2+ and EDTA present in the solution:
Moles of Ni2+ (n1) = C1 * V1 = 0.0300 M * 0.0500 L
Moles of EDTA (n2) = C2 * V2 = 0.0500 M * 0.0500 L
Now, we need to determine the concentration of Y4- in the solution. At pH = 3, we know that α4 = 2.5x10^-11 and KNiY = 4.2x10^18.
The equation for α4 is given as:
α4 = [NiY2-] / [Ni2+] * [Y4-]^4
We can rearrange the equation to solve for [Y4-]:
[Y4-] = (α4 * [Ni2+] / [NiY2-])^(1/4)
Substituting the given values:
[Y4-] = (2.5x10^-11 * 0.0300 M) / (4.2x10^18)^(1/4)
Now, we can substitute the value of [Y4-] into the complexation equilibrium equation:
[Ni2+] = [NiY2-] / [Y4-]
[Ni2+] = (0.0300 M) / [(2.5x10^-11 * 0.0300 M) / (4.2x10^18)^(1/4)]
Solving the equation will give you the concentration of Ni2+ in the solution.
Note:
Please make sure to double-check the calculations and units to ensure accuracy.