Question

LAB DESCRIPTION:

In aqueous solutions the Ni2+ cation forms a complex ion with 6 H2O ligands. The formula of this complex ion is: Ni(H2O)62+. When this complex ion is mixed with aqueous NH3 a reaction occurs where some of the H2O ligands are displaced with NH3 ligands. But how many of the H2O are displaced by NH3?

In this experiment you will mix nickel(II) sulfate and aqueous ammonia to synthesize a compound containing a complex ion whose formula is Ni(H2O)6-x(NH3)x2+, where x= the number of H2O ligands that were displaced by NH3 ligands. It is your job in this lab to determine the value of x and hence the formula of this complex ion. You can do this by performing some simple titrations on samples of this compound. First you will analyze the compound to determine the number of moles of NH3 present per gram of compound. Then you will find the number of moles of Ni2+ per gram. The fraction:

molesNH3/gmolesNH3 x moles Ni2 / g moles Ni2

Since NH3 acts as a base in aqueous solution, you can analyze the compound for NH3 content using an acid-base titration. You will treat a portion of the compound with a known volume of HCl solution of known molarity. This delivers a known number of moles of HCl. Some of this HCl is neutralized by the NH3:

HCl + NH3  NH4Cl.
The amount of HCl remaining is determined by titrating the mixture with standardized NaOH:

NaOH + HCl  NaCl + H2O.

The amount of NH3 present is then found by realizing that all of the HCl must have been neutralized by some base, either NH3 or NaOH. Whatever NaOH did not neutralize must have been neutralized by NH3. So the of mol HCl taken = mol NH3 in compound + mol NaOH added. You can easily find the number of moles of NH3 = mol HCl – mol NaOH. This type of analysis is called an indirect titration.

The Ni2+ ion, as you have seen, acts as a Lewis acid with the Lewis base NH3 to form the complex ion. Ni2+ forms a stronger complex ion with the hexadentate ligand, EDTA4-. You can determine the amount of Ni2+ in your compound by titrating a sample of it with an EDTA4- solution:

Ni(H2O)6-x(NH3)x2+ (aq) + EDTA4-(aq)  NiEDTA2- (aq) + x NH3 (aq) + (6-x) H2O

A difficulty with this titration is the fact that both complex ions, Ni(H2O)6-x(NH3)x2+, and NiEDTA2- are blue in solution. This makes it impossible to tell when the titration is finished. This problem is solved by the use of an indicator called murexide. Murexide is another ligand

1

that is red by itself and forms a golden yellow colored complex ion with Ni2+. You will dissolve a sample of the compound in a buffered aqueous solution. The addition of murexide makes the solution a golden yellow color. You will then add EDTA solution from a buret until the yellow color is gone and the solution becomes purple.

QUESTION:

1. If the sample of the complex ion compound that you prepared had not thoroughly dried, but contained traces of water or ethanol would this make your determination of the moles NH3/g and moles Ni2+/g too high, too low, or would it not affect your results at all? Would this make your determination of the value of x too high, too low or would it not affect your result at all? Explain

Answer 1

The moles NH3/g too low.

The moles Ni2+/g too high.

The value of 'x' would not get affected.

Explanation: The overall moles of NH3 to be reacted with added acid (HCl) would not change, as a result, the value of 'x' will remain the same.