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

The coefficients of reactants tell us that 1 mole of Cu reacts per 2 moles of AgNO₃. We want to know the mass of Cu that will react with a mass (specifically, 420 g) of AgNO₃. To start, we must convert the mass of AgNO₃ to moles of AgNO₃:

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Moles of AgNO₃ = 420 g AgNO₃ ÷ 169.88 g AgNO₃/mol AgNO₃

Moles of AgNO₃ = 2.47 moles of AgNO₃.

Since only half as many moles of Cu is consumed in reacting with the AgNO₃, the number of moles of Cu that will react with 2.47 moles of AgNO₃ would be half of 2.47, or 1.24 moles of Cu.

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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.

So, as we did with the first question, we first convert the mass of the Ag produced to moles. The molar mass of Ag is 107.8682 g/mol. Dividing 98.5 g of Ag by the molar mass gives us 0.913 moles of Ag produced. The number of moles of Cu that reacted to produce 0.913 moles of Ag would be 0.913/2 = 0.457 moles of Cu.

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