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Lubov Fominskaja [6]
3 years ago
5

EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK

a3=2.00 , p K a4 = 2.69 pKa4=2.69 , p K a5 = 6.13 pKa5=6.13 , and p K a6 = 10.37 pKa6=10.37 . The distribution of the various protonated forms of EDTA will therefore vary with pH. For equilibrium calculations involving metal complexes with EDTA EDTA , it is convenient to calculate the fraction of EDTA EDTA that is in the completely unprotonated form, Y 4 − Y4− . This fraction is designated α Y 4 − αY4− . Calculate α Y 4 − αY4− at two pH values.
Chemistry
1 answer:
mihalych1998 [28]3 years ago
4 0

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = 5.0 X 10^{-4} M

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}

alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6

= 1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}

= 2.02 X 10^{-13}

When,

pH = -log[H+] = 10.15

[H+] = 7.08 X 10^{-11} M

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^-11

alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6

= 1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}

= 5.12 X 10^{-23}

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choose any bases that will selectively deprotonate the above acid, that is, any bases that will favor the formation of products.
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In order to deprotonate an acid, we must remove protons in order to achieve a more stable conjugate base. For this example, we can use the relationship between carboxylic acid and hydroxide.

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8 0
2 years ago
How many moles of water are in a beaker with 50 mL?
tatiyna

Answer:

Number of moles = 2.8 mol

Explanation:

Given data:

Number of moles of water = ?

Volume of water = 50 mL

Density of water = 1.00 g/cm³

Solution:

1 cm³ =  1 mL

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