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OLga [1]
3 years ago
10

A student has a 0.00124 M HCl (aq) solution and she titrates 25.00 mL of this solution against an unknown potassium hydroxide so

lution. The acid requires 44.25 mL of base to reach the end point, what is the molarity of the KOH (aq) solution?
Chemistry
1 answer:
malfutka [58]3 years ago
5 0

Answer:

The Molarity of KOH is

7,01.10^{-4}M

Explanation:

The endpoint indicates the volume necessary to neutralize the moles of acid.

In other words, the point at which the moles of both solutions are the same.

M_{(HCl)}V_{HCl}=n\\ \\M_{(KOH)}V_{KOH}=n

we match these equations and find the concentration of KOH

M_{(HCl)}. V_{(HCl)} =M_{(KOH)} .V_{(KOH)}\\ \\M_{(KOH)}=\frac{M_{(HCl)}. V_{(HCl)}}{V_{(KOH)}} \\\\M_{(KOH) =\frac{(25mL)(0,00124m)}{(44,25mL)}\\\\

M_{(KOH)}=7,01.10^{-4}

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Consider the following system at equilibrium:A(aq)+B(aq) &lt;---&gt; 2C(aq)Classify each of the following actions by whether it
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Explanation:

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This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

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  • On addition of product at equilibrium shifts the equilibrium in backward direction.
  • On removal of reactant at equilibrium shifts the equilibrium in backward direction.
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A(aq)+B(aq)\rightleftharpoons 2C(aq)

Reactants = A , B

Product = C

1. Increase A

On increasing the amount of A at equilibrium will shift the equilibrium in forward or rightward direction.

2. Increase B

On increasing the amount of B at equilibrium will shift the equilibrium in forward or rightward direction.

3. Increase C

On increasing the amount of C at equilibrium will shift the equilibrium in backward or leftward direction.

4. Decease A

On decreasing the amount of A at equilibrium will shift the equilibrium in backward or leftward direction.

5. Decease B

On decreasing the amount of B at equilibrium will shift the equilibrium in backward or leftward direction.

6. Decease C

On decreasing the amount of C at equilibrium will shift the equilibrium in forward or rightward direction.

7. Double A and Halve B

Equilibrium constant of the reaction = K

K=\frac{[C]^2}{[A][B]}

On doubling A and halving B, equilibrium constant of the reaction = K'

K'=\frac{[C]^2}{[2A][\frac{B}{2}]}=\frac{[C]^2}{[A][B]}

The value of equilibrium constant K' is equal to K, which means that equilibrium will not shift in any direction.

8. Double both B and C

Equilibrium constant of the reaction = K

K=\frac{[C]^2}{[A][B]}

On doubling B and C, equilibrium constant of the reaction = K'

K'=\frac{[2C]^2}{[A][2B]}=\frac{4[C]^2}{[A][2B]}=\frac{2[C]^2}{[A][B]}

K' = 2 K

The value of equilibrium constant K' is double the K, which means that product is increasing which means that equilibrium will shift in backward or leftward direction.

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Answer:

Explanation:

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