Question

In: Chemistry

A 130.0 mL sample of 0.070 M Ca 2 + is titrated with 0.070 M EDTA...

A 130.0 mL sample of 0.070 M Ca 2 + is titrated with 0.070 M EDTA at pH 9.00. The value of log K f for the Ca 2 + − EDTA complex is 10.65 and the fraction of free EDTA in the Y 4 − form, α Y 4 − , is 0.041 at pH 9.00. What is K ′ f , the conditional formation constant, for Ca 2 + at pH 9.00?

What is the equivalence point volume, Ve, in milliliters?

Calculate the concentration of Ca2+ at V=(1/2)Ve.

Calculate the concentration of Ca2+ at V=Ve.

Calculate the concentration of Ca2+ at V=1.1Ve.

Solutions

Expert Solution


Related Solutions

A 100.00 mL sample of hard water is to be titrated with 0.03498 M EDTA solution....
A 100.00 mL sample of hard water is to be titrated with 0.03498 M EDTA solution. A small amount of Mg<sup>2+</sup> (that required 2.52 mL of EDTA to titrate as a blank solution) is added to the hard water sample. This sample requires 17.89 mL of EDTA to reach the endpoint. How many moles of Ca<sup>2+</sup> are in the hard water? Express answer as a decimal, not an exponent.
?. C3) A 100.00 ml. water sample was titrated 20.50 mL of 0.150 M EDTA after...
?. C3) A 100.00 ml. water sample was titrated 20.50 mL of 0.150 M EDTA after pH adjustment and using Eriochrome black T indicator. Knowing that density of water sample equals to 1.10 g/mL. (Caco 00.09 amu). Calculate the total hardness of water as Cacos and accordingly classify
A 25.00 ml sample of Ca(OH)2 (aq) is titrated with 0.05668 M HCL. The titration requires...
A 25.00 ml sample of Ca(OH)2 (aq) is titrated with 0.05668 M HCL. The titration requires 18.88 ml of HCL (aq) to teach the endpoint. 1) Write a net-ionic equation for the titration reaction. 2) Determine the molarity of the Ca(OH)2 (aq) 3) Express the molarity of the CA(OH)(aq) as a solubility in g Ca(OH)2/ 100 ml soln.
3)A 100 mL sample of hard water is titrated with 22.4 mL of EDTA solution from...
3)A 100 mL sample of hard water is titrated with 22.4 mL of EDTA solution from problem 2. the same amount of MgCl2 is added as previously and the total volume of EDTA solution required is 22.44 mL. A)What volume of EDTA is used titration the Ca2+ in the hard water? B)how many moles of EDTA are there in the volume? C)how many moles of Ca2+ are there in 100 mL of water? D)assume all of the Ca2+ I'm the...
A 130.0 mL sample of a solution that is 0.0126 M in NiCl2 is mixed with...
A 130.0 mL sample of a solution that is 0.0126 M in NiCl2 is mixed with a 190.0 mL sample of a solution that is 0.400 M in NH3. -After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108.
A 25.00 mL solution containing 0.3000 M Na2C2O4 (sodium oxalate) was titrated with 0.1500 M Ca(NO3)2...
A 25.00 mL solution containing 0.3000 M Na2C2O4 (sodium oxalate) was titrated with 0.1500 M Ca(NO3)2 to precipitate calcium oxalate: Use Excel to draw a graph of pCa vs. VCa from 0 to 75 mL, using a suitable resolution to show a sharp endpoint. Label the axes and show where the end point lies.
2) A 25.0 mL sample of 0.160 M KOH is titrated with 0.250 M HCl. Determine...
2) A 25.0 mL sample of 0.160 M KOH is titrated with 0.250 M HCl. Determine the following quantities: a) The initial pH of the KOH solution before any acid is added. b) The pH of the solution after 3.00 mL of HCl is added. c) The quantity of HCl solution required to reach the equivalence point. d) The pH of the solution at the equivalence point. e) The pH of the solution after20.0 mL of HCl is added.
A 35 mL sample of water is triated with 0.0100 M EDTA. Exactly 9.70 mL of...
A 35 mL sample of water is triated with 0.0100 M EDTA. Exactly 9.70 mL of EDTA are required to reach the EBT endpoint. Calculate the total hardness in ppm CaCO3.
A 20.00 mL sample of 0.120 M NaOH was titrated with 52.00 mL of 0.175 M...
A 20.00 mL sample of 0.120 M NaOH was titrated with 52.00 mL of 0.175 M HNO3.             Calculate the [H+], [OH-], pH, and pOH for the resulting solution.   
A 130.0 mL sample of a solution that is 2.7×10−3 M in AgNO3 is mixed with...
A 130.0 mL sample of a solution that is 2.7×10−3 M in AgNO3 is mixed with a 230.0 mL sample of a solution that is 0.14 M in NaCN. After the solution reaches equilibrium, what concentration of Ag+(aq) remains?
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT