Question

In: Chemistry

An aliquot containing 0.0004693 moles of Ca2+ is titrated to the Eriochrome Black T end point....

An aliquot containing 0.0004693 moles of Ca2+ is titrated to the Eriochrome Black T end point. A blank containing a small amount of measured Mg2+requires 2.91 mL of EDTA to reach the end point. The aliquot to which the same amount of Mg2+ is added requires 25.59 mL of EDTA solution to reach the end point. What is the molarity of the EDTA solution? Express as a decimal number, not an exponent.

Solutions

Expert Solution

Ca2+ ion and EDTA reacts 1:1 mole ratio.

So, the number of moles of EDTA required to titrate 0.0004693 moles of Ca2+ is 0.0004693 moles.

mL of EDTA consumed for Ca2+ ion = mL EDTA needed for sample - mL EDTA consumed for blank

                                                           = 25.59 mL - 2.91 mL

                                                           = 22.68 mL

Since 1 L = 1000 mL, 22.68 mL = 0.02268 L.

Molarity of EDTA = moles of EDTA/volume of EDTA = 0.0004693 mol/0.02268 L = 0.02069 mol/L = 0.02069 M.


Related Solutions

A beaker containing 50.0 ml of 0.300M Ca2+ at ph 9 is titrated with 0.150M EDTA....
A beaker containing 50.0 ml of 0.300M Ca2+ at ph 9 is titrated with 0.150M EDTA. Calculate the pCa2+ at the equivalence point
A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence...
A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence point is reached after 21.27 mL of the base were added. Calculate 1) the concentration of the acid in the original solution, 2) the pH of the original HCl solution and the original NaOH solution, 3) the pH after 10.00 mL of NaOH have been added, 4) the pH at the equivalence point, and 5) the pH after 25.00 mL of NaOH have been...
A 1.000 mL aliquot of a solution containing Cu 2 + and Ni 2 + is...
A 1.000 mL aliquot of a solution containing Cu 2 + and Ni 2 + is treated with 25.00 mL of a 0.05485 M EDTA solution. The solution is then back titrated with 0.02234 M Zn 2 + solution at a pH of 5. A volume of 21.46 mL of the Zn 2 + solution was needed to reach the xylenol orange end point. A 2.000 mL aliquot of the Cu 2 + and Ni 2 + solution is fed...
A 40.00 mL aliquot of 0.100 M NH3 is titrated with 20.00 mL of 0.0700 M...
A 40.00 mL aliquot of 0.100 M NH3 is titrated with 20.00 mL of 0.0700 M HCl. Calculate the pH. Kb = 1.76 x 10−5
A 50.0ml aliquot of 0.100 M NaOH is titrated with 0.200 M HCl. Calculate the pH...
A 50.0ml aliquot of 0.100 M NaOH is titrated with 0.200 M HCl. Calculate the pH of the solution after addition of 0.0, 5.0, 10.0, 25.0, 35.0, 45.0, 55.0 of HCl. Plot the titration curve in a graph paper.
Consider the titration of a 50.00 mL aliquot of a water sample containing Fe3++ and Pb2++...
Consider the titration of a 50.00 mL aliquot of a water sample containing Fe3++ and Pb2++ with 19.62 mL of 0.00243 M EDTA to reach the end point. Excess CN- was added to a separate 20.00 mL aliquot of the same water sample, causing Fe(CN)3 to precipitate. This aliquot was titrated with 2.01 mL of 0.00243 M EDTA to reach the end point. What is the molarity of Fe3+ in the water sample? Please explain every step. Thank you.
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.03742 M EDTA solution. The solution is then back titrated with 0.02190 M Zn2 solution at a pH of 5. A volume of 20.48 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.05893 M EDTA solution. The solution is then back titrated with 0.02306 M Zn2 solution at a pH of 5. A volume of 19.89 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.04503 M EDTA solution. The solution is then back titrated with 0.02327 M Zn2 solution at a pH of 5. A volume of 22.80 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
A 110.0-mL aliquot of 0.120 M weak base B (pKb = 4.55) was titrated with 1.20...
A 110.0-mL aliquot of 0.120 M weak base B (pKb = 4.55) was titrated with 1.20 M HClO4. Find the pH at the following volumes of acid added: Va = 0.00, 1.30, 5.50, 10.00, 10.90, 11.00, 11.10, and 14.00 mL. (Assume Kw = 1.01 ✕ 10−14.)
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT