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Readme [11.4K]
3 years ago
15

Can yall help me with this please?

Chemistry
2 answers:
Serhud [2]3 years ago
7 0
Hi i think its the the 3rd one 
adoni [48]3 years ago
7 0
Each of the first three options would decrease the final yield of reaction in some way, while a higher percent yield results in a higher actual yield if the same amount of reactant is produced, therefore the answer is the 4th option.
Hope this helps!
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A 1-liter solution contains 0.494 M hydrofluoric acid and 0.371 M potassium fluoride. Addition of 0.408 moles of hydrochloric ac
UkoKoshka [18]

Answer:

Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

Explanation:

The pH of the buffer solution before the addition of HCl is:

pH = pKa + log(\frac{[KF]}{[HF]})

pH = -log(6.8 \cdot 10^{-4}) + log(\frac{0.371}{0.494}) = 3.04  

The hydrochloric acid added will react with the potassium fluoride as follows:

H₃O⁺(aq)  +  F⁻(aq) ⇄   HF(aq) + H₂O(l)

The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:

\eta_{KF}_{i} = C_{KF}*V = 0.371 M*1 L = 0.371 mol

\eta_{HF}_{i} = C_{HF}*V = 0.494 M*1 L = 0.494 moles

The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:

\eta_{HCl} = \eta_{HCl} - \eta_{KF}_{i} = 0.408 moles - 0.371 moles = 0.037 moles  

Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.  

We can calculate the pH after the addition of HCl:

HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq)    (1)

The number of moles of HF after the reaction of KF with HCl is:

\eta_{HF} = 0.494 moles + (0.408 moles - 0.371 moles) = 0.531 moles

And the concentration of HF after the reaction of KF with HCl is is:

C_{HF} = \frac{\eta_{HF}}{V} = \frac{0.531 moles}{1 L} = 0.531 moles/L

Now, from the equilibrium of equation (1) we have:

Ka = \frac{[H_{3}O^{+}][F^{-}]}{[HF]}

Ka = \frac{x^{2}}{0.531 - x}  (2)

By solving equation (2) for x we have:

x = 0.0187

Finally, the pH after the addition of HCl is:

pH = -log (H_{3}O^{+}) = -log (0.0187) = 1.73

Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

I hope it helps you!

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4 years ago
Please help with this chemistry question!
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Hey there!

Compounds with ionic bonds have higher melting points because of the forces needed to break through the strong forces of attraction holding it together.

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So, your answer is C. Compound 1 is ionic, and compound 2 is molecular.

Hope this helps!

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