In: Chemistry

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

Consider a complex formation reaction: Ca ^{2+} + EDTA
[ CaY ]
^{2-}

For above reaction, conditional formation constant is K'
_{f} = (_{Y}
4-) K _{f}

Therefore, K' _{f} = 0.041 x 10 ^{10.65} = 1.83
x 10 ^{9}

First , we Calculate of volume of EDTA required to reach equivalence point

From above reaction , we have 1 mol Ca ^{2+} 1
mol EDTA

Therefore, M _{ca} 2+ V
_{Ca}2+ = M _{EDTA} V
_{EDTA}

V
_{EDTA} = M _{ca} 2+ V
_{Ca}2+ / M _{EDTA} =

V _{EDTA} = 0.300 M 50.0 ml /
0.150 M

V _{EDTA} = 100 ml

volume of EDTA required to reach equivalence point = 100 ml

At the equivalence point, all the calcium ions are consumed by
added EDTA. Hence, at this stage concentration of Ca ^{2+}
will be due to dissociation of [CaY] ^{2-}.

Moles of [ CaY ] ^{2-} = moles of EDTA added to the
solution

No. of moles of EDTA added in the solution = 0.150 mol / L 0.100 L = 0.015 mol

Volume of solution at equivalence point = Volume of Ca
^{2+} solution + volume of EDTA solution

Volume of solution at equivalence point = 50.0 + 100 ml =150.0 ml = 0.150 L

[CaY] ^{2-}. = 0.015 mol / 0.150 L = 0.100 M

Let’s use ICE table.

Concentration ( M ) | Ca ^{2+} |
EDTA | [CaY] ^{2-} |

I | 0.100 | ||

C | +X | +X | -X |

E | X | X | 0.100 - X |

We have , K' _{f} = [CaY^{2-}] / [Ca
^{2+}] [EDTA]

Therefore,K' _{f} = 0.100 -X / (X)(X) = 1.8310
^{9}

0.100 -X / (X)(X) = 1.83 10
^{9}

0.100 -X / X ^{2} = 1.83 10
^{9}

At equilibrium X is negligible as compared 0.100. Then we can
write 0.100 / x ^{2} = 1.83 x 10 ^{9}

X
^{2} = 0.100 / 1.83 10
^{9} =5.464 10
^{-11}

X = 7.392
10 ^{-
6} M

[Ca
^{2+}] = 7.932 10 ^{-
6} M

We have relation, pCa ^{2+} = - log [Ca
^{2+}]

pCa
^{2+} = - log 7.932 10 ^{-
6} = 5.13

**ANSWER : pCa ^{2+} at equivalence point =
5.13**

A 50.0-mL solution containing Ni2+ and Zn2+ was
treated with 25.0 mL of 0.0452 M EDTA to bind all the metal. The
excess unreated EDTA required 12.4 mL of 0.0123 M Mg2+ for complete
reaction. An excess of the reagent 2,3-dimercapto-1-propanol was
then added to displace the EDTA from zinc. Another 29.2 mL of Mg2+
was required for reaction with the liberated EDTA. Calculate the
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You are titrating 110.0 mL of 0.050 M Ca2 with 0.050 M EDTA at
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PLEASE ANSWER ALL PARTS: A, B, C, D, AND E.
(a) What is K\'f, the conditional formation constant, for Ca2 at
pH 9.00?
(b) What is the equivalence volume, Ve, in milliliters?
(c) Calculate the concentration of Ca2 at...

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(b) What is the equivalence volume, Ve, in milliliters?
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?.
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Calculate the total hardness of water as Cacos and accordingly
classify

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An aliquot containing 0.0004693 moles of Ca2+ is
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reach the end point. What is the molarity of the EDTA solution?
Express as a decimal number, not an exponent.

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How many moles of Ca<sup>2+</sup> are in the hard
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Express answer as a decimal, not an exponent.

You have 3 beakers:
Beaker 1 contains 50.0 ml of 0.123 M
HC2H3O2
Beaker 2 contains 25.0 ml of 0.456 M
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A.) What is the pH of the solution in beaker
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in step 32 a student added 150 mL of water to the beaker
containing 10 mL of unknown metal nitrate solution instead of 100
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justify your answer

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