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3 years ago

Aqueous copper (II) sulfate reacts with aqueous potassium fluoride to produce

Chemistry

1 answer:

Nina [5.8K]3 years ago

8 0

Answer:

CuSO₄(aq) + 2 KF(aq) = CuF₂ + K₂SO₄

Explanation:

The question is missing but I think it must be about writing and balancing the equation.

Let's consider the unbalanced equation for the reaction that occurs when aqueous copper (II) sulfate reacts with aqueous potassium fluoride to produce a precipitate of copper (II) fluoride (I fixed a mistake here) and aqueous potassium sulfate. This is a double displacement reaction.

CuSO₄(aq) + KF(aq) = CuF₂ + K₂SO₄

Since only K and F atoms are not balanced, we will get the balanced equation by multiplying KF by 2.

CuSO₄(aq) + 2 KF(aq) = CuF₂ + K₂SO₄

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Read 2 more answers

Calculate the pH of the following simple solutions:

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Answer :

(1) pH = 1.27

(2) pH = 13.35

(3) The given solution is not a buffer.

Explanation :

(1) 53.1 mM HCl

Concentration of HCl = $53.1mM=53.1\times 10^{-3}M$

As HCl is a strong acid. So, it dissociates completely to give hydrogen ion and chloride ion.

So, Concentration of hydrogen ion= $53.1\times 10^{-3}M$

pH : It is defined as the negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pH=-\log (53.1\times 10^{-3})$

$pH=1.27$

(2) 0.223 M KOH

Concentration of KOH = 0.223 M

As KOH is a strong base. So, it dissociates completely to give hydroxide ion and potassium ion.

So, Concentration of hydroxide ion= 0.223 M

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (0.223)$

$pOH=0.65$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-0.65=13.35$

(3) 53.1 mM HCl + 0.223 M KOH

Buffer : It is defined as a solution that maintain the pH of the solution by adding the small amount of acid or a base.

It is not a buffer because HCl is a strong acid and KOH is a strong base. Both dissociates completely.

As we know that the pH of strong acid and strong base solution is always 7.

So, the given solution is not a buffer.

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Answer:

Explanation:

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