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Andreyy89
3 years ago
15

Calculate the ph of a 0.40 m solution of sodium benzoate (nac6h5coo) given that the ka of benzoic acid (c6h5cooh) is 6.50 x 10-5

.
a. 2.29
b. 9.81
c. 5.11
d. 8.89
e. 11.71
Chemistry
1 answer:
asambeis [7]3 years ago
3 0
Hello!

The dissociation reaction for Benzoic Acid is the following:

C₆H₅COOH + H₂O ⇄ C₆H₅COO⁻ + H₃O⁺

The Ka expression is the following and we clear for the concentration of H₃O⁺(X) assuming that the dissociation is little so we can rule it out in the denominator of the equation:

Ka= \frac{[C_6H_5COO^{-}]*[H_3O^{+}] }{[C_6H_5COOH]}=\frac{X*X }{0,40 -X}  (assume: 0,40-X\approx0,40)\\  \\ X= \sqrt{0,40*6,50*10^{-5} }=0.00510M \\  \\ pH=-log([H_3O^{+}]=2,29

So, the pH of this Benzoic Acid solution is 2,29

Have a nice day!


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Answer:

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Given ,  

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T₂ = 253 K

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\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}

\frac{{3.39}\times {25.0}}{353}=\frac{{1.35}\times {V_2}}{253}

\frac{1.35V_2}{253}=\frac{3.39\times \:25}{353}

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T or F in an argument, but not alone. Can be a premise or conclusion. Is not equal to a sentence.

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