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
molarity of Ni2+ and Zn2+ in the original solution

You are titrating 110.0 mL of 0.050 M Ca2 with 0.050 M EDTA at
pH 9.00. Log Kf for the Ca2 -EDTA complex is 10.65, and the
fraction of free EDTA in the Y4– form, αY4–, is 0.041 at pH 9.00.
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...

You are titrating 120.0 mL of 0.080 M Ca2+ with 0.080 M EDTA at
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9.00.
(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 V = 1/2 Ve.
(d) Calculate the concentration of Ca2...

Determine the pH at the equivalence point when 0.05 M KOH is
titrated with 50.0 ml of 0.01 M benzoic acid
HC7H5O2. The Ka for
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A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH.
Calculate the pH after the addition of 13.0 mL of KOH. Express your
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A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5)
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Express your answer numerically.
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Calculate the pH after the addition of 33.0 mL...

?.
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

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
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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 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
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Express answer as a decimal, not an exponent.

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titrated to the Eriochrome Black T end point. A blank containing a
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You have 3 beakers:
Beaker 1 contains 50.0 ml of 0.123 M
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Beaker 2 contains 25.0 ml of 0.456 M
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Beaker 3 contains 5.0 ml of 0.789 M HCl
A.) What is the pH of the solution in beaker
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B.) The contents of beaker 1 are poured into Beaker 2
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C.) The contents of beaker 3 are now poured into the
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