Question

2. Calculate the concentration of one liter of the resulting calcium chloride solution (assuming you started with 102.7 mg of calcium carbonate) in units of molar and millimolar. Also calculate the concentration in “parts-per-million as CaCO3”. (Refer to section 13.5 of your textbook, if needed.)

a. If a standard Ca2+ solution is 100 ppm and 50 mL of solution is used for the standardization, calculate the approximate volume of 2.5 mM EDTA4- solution needed to reach the end point of the titration.

The endpoint of an EDTA titration is determined with a metallochromic indicator. These indicators are complexing agents that change color when combined with metal ions. A variety of indicators can be used for EDTA titrations. In this experiment, we will use "Eriochrome black T" (EBT) indicator, which has the structure shown below: 1 Na = 0 NEN HO This indicator (shown as the monoanion H2In in the equations below) changes from blue to red when combined with a metal ion such as Ca2+, forming a complex ion: Ca? + H2In" + 2H2O (BLUE) Caln" + 2H30 (RED) Since EDTA is a stronger complexing agent than the indicator, it displaces the indicator from the metal ion, allowing the indicator to return (through shades of violet) to a pure blue color, indicating the endpoint of the reaction.

b. As you can see in the structure (above, top of page 4), the indicator EBT has two dissociable protons (H+) on hydroxide groups (-OH); they are weak acids and both of their pKa values are around ~10. Likewise, EDTA has four dissociable protons and the associated values for pKa are 2.0, 2.7. 6.2, and 10.3. Which protons are more acidic – those on EBT or those on EDTA? This experiment will be conducted in buffer near pH = 10. Are the protons for EBT mostly dissociated or undissociated at this pH? What about the protons for EDTA?

Answer 1

2. Calculate the concentration of one liter of the resulting calcium chloride solution (assuming you started with 102.7 mg of calcium carbonate) in units of molar and millimolar. Also calculate the concentration in “parts-per-million as CaCO3”.

moles of CaCO3 : 102.7 mg ==> 0.1027 g

thus, = 0.1027 g / 100 g/mol = 0.001027 moles

Since 1 moles of CaCO3 = 1 mol CaCl2

thus, [CaCl2] = 0.001027 mol/L = 0.001027 M

a. If a standard Ca2+ solution is 100 ppm and 50 mL of solution is used for the standardization, calculate the approximate volume of 2.5 mM EDTA4- solution needed to reach the end point of the titration.

Ans.

Molarity of 100 ppm Solution : 100 mg/L ==> 0.1 g/L

0.1g/L / 100 g/mol * 1000 mmol/1mol = 1 mM

thus 1 mM solution

EDTA titration results in formation of complexes with 1:1 to ration with metal ion:

Ca2+ + Y4- → Ca- Y2-

V(EDTA) = 50 mL*1 mM / 2.5 mM = 20 mL EDTA solution.

b. As you can see in the structure , the indicator EBT has two dissociable protons (H+) on hydroxide groups (-OH); they are weak acids and both of their pKa values are around ~10. Likewise, EDTA has four dissociable protons and the associated values for pKa are 2.0, 2.7. 6.2, and 10.3. Which protons are more acidic – those on EBT or those on EDTA? This experiment will be conducted in buffer near pH = 10. Are the protons for EBT mostly dissociated or undissociated at this pH? What about the protons for EDTA?

Since pKa of EDTA protons is lesser than that of EBT, pKa = -log Ka

thus Higher Ka,thus EDTA protons are more acidic than those on EBT .

at pH 10 , we take EBT pKa ~ 10,

than from, pH = pKa+ log[In- /HIn]

we can predict neary Half potons for first pKa will dissociated.

= Almost all protons of EDTA will be dissociated.