{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

prelab2 - CaCO 3 solution that will be prepared After the...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Pre-lab experiment #2 Hard water Introduction: Water contains dissolved ions, minerals, and metals including Ca 2+ and Mg 2+ which contribute to the “hardness” of water. Soaps, which contain singly-charged K + and Na + ions, are replaced by the heavy metals, ions, and minerals, forming a precipitate called scale when heated and acting as an insulator. The scale and precipitate build up causing clogs and rings in tubs. Calcium and Magnesium ions are the most common hard water ions, therefore the concentration of Ca 2+ plus the concentration of Mg 2+ ions is equal to the hardness of water. By titrating the water (with a metal bonding indicator) with a tetraprotic chelating agent called EDTA, the concentration of Calcium and Magnesium ions can be calculated, yielding the hardness of the water expressed in CaCO 3 mg/L or ppm. Procedure Summary: First, a solution of EDTA will be prepared by diluting about one gram of EDTA with 250 mL of distilled water. This solution will then be standardized by titration three times against a standard
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CaCO 3 solution that will be prepared. After the molarity of the EDTA solution is calculated and standardized, a sample of CaCO 3 with an unknown molar concentration will be obtained and titrated three times with the EDTA solution to the end point color change from wine to blue. The volume of titrant added will be recorded and the molarity of the unknown solution can be calculated form this data. Chemical reactions: Ca 2+ + Y 4----------> CaY 2-K f = [CaY 2-] / [Ca 2+ ][Y 4-] = 10 10.69 Mg 2+ + Y 4----------> MgY 2-K f = [MgY 2-] / [Mg 2+ ][Y 4-] = 10 8.79 Questions and calculations: If you have 25mL of a standard 1 mg/mL CaCO 3 solution how many mLs of the 0.010 M solution do you expect to use in the titration? 25 mL X 1mg/1mL X 1mmol/100mg X 1mol CaCO 3 /1000mmol X 1mol EDTA/1mol CaCO 3 X 1000mL/0.010mol EDTA=25mL of 0.010 mol EDTA solution...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online