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Calculate the concentrations of all species in a 1.17 M NaCH3COO (sodium acetate) solution. The ionization constant for acetic acid is a=1.8×10−5 .

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okpalawalter8

The concentrations of all species in a 1.17 M NaCH3COO solution is mathematically given as

[Na+] = 1.17 M

[OH-] = 0.00002559 M

[H3O+] = 3.51 x 10-10 M

[CH3COO-] = 1.17 M

[CH3COOH] = 0.00002559 M

What are the concentrations of all species in a 1.17 M NaCH3COO (sodium acetate) solution.?

Question Parameters:

1.17 M NaCH3COO (sodium acetate) solution

The ionization constant for acetic acid is a=1.8×10−5.

Generally, the equation for the Chemical Reaction   is mathematically given as

NaCH2COO - +H20---><--- CH3COOH+OH-

Therefore

[tex]Ka=\frac{cH3COOH(OH)}{CH2OO-}[/tex]

[tex]5.6*10^{-10}=\frac{*x*x}{1.17m-x}[/tex]

x=0.00002559

In conclusion,

Since other species are given by either for x=0.00002559 or the ionization constant

For[CH3COO-]

[CH3COO-]=1.17-0.00002559

[CH3COO-]=1.167m

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