Benzoic acid, HC6H5CO2,is a monoprotic acid(only one H+ ionizes)with a Ka=6.5×10^-5. Calculate [H+] and the pH of a .32M solution of benzoic acid. PLEASE ANSWER. ​

Answers

Answer 1

Answer:

[H⁺] = 4.56x10⁻³ M

pH = 2.34

Explanation:

Hello there!

In this case, according to the ionization reaction of benzoic acid:

[tex]HC_6H_5CO_2+H_2O\rightleftharpoons C_6H_5CO_2^++H_3O^+[/tex]

Whereas [tex][H_3O^+]=[H^+][/tex], we can set up the equilibrium expression in terms of [tex]x[/tex] (reaction extent) to obtain:

[tex]Ka=\frac{[C_6H_5CO_2^-][H_3O^+]}{[HC_6H_5CO_2]} \\\\6.5x10^{-5}=\frac{x^2}{0.32-x}[/tex]

However, since Ka<<<1, we can neglect the [tex]x[/tex] on bottom to easily solve for it:

[tex]6.5x10^{-5}=\frac{x^2}{0.32}\\\\x=\sqrt{6.5x10^{-5}*0.32} \\\\x=4.56x10^{-3}[/tex]

Which is actually the same as [H⁺]. Finally, the pH turns out to be:

[tex]pH=-log(4.56x10^{-3})\\\\pH=2.34[/tex]

Best regards!


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Answer:

You can't tell with the information provided.

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Since this is a balanced chemical equation, there is no limiting reactant. Can you tell us how much of the reactants you began with?

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Answers

Answer:

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Answers

Answer:

The initial volume of the balloon is 3.41 L.

Explanation:

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Boyle's law is expressed mathematically as:

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In this case:

P1= 0.53 atmV1= ?P2= 0.42 atmV2= 4.3 L

Replacing:

0.53 atm* V1= 0.42 atm* 4.3 L

Solving:

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V1= 3.41 L

The initial volume of the balloon is 3.41 L.

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Answer:

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Answers

Answer:

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Answer:D

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Answers

Answer:

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Explanation:

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Right on Edge

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Answers

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brainly.com/question/16031275

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Answers

Answer:

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Answer:

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HOPE THIS HELPS

HAVE A GOOD DAY

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Our balanced equation is Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag.

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---

The way to solve the second question is identical to how we solved the first question. The coefficient of Ag is 2, so the molar ratio between the Ag produced and Cu reacted is 2:1. In other words, half as many moles of Ag produced is equivalent to the number of moles of Cu that was consumed in the reaction.

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Answers

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Answers

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Answer:

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