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3 years ago

The pOH of a solution of KOH is 11.30. What is the [H*] for this solution?

Chemistry

1 answer:

Sloan [31]3 years ago

4 0

Answer: The concentration of hydrogen ions for this solution is $1.99 \times 10^{-3}$ .

Explanation:

Given: pOH = 11.30

The relation between pH and pOH is as follows.

pH + pOH = 14

pH + 11.30 = 14

pH = 14 - 11.30

= 2.7

Also, pH is the negative logarithm of concentration of hydrogen ions.

$pH = - log [H^{+}]$

Substitute the values into above formula as follows.

$pH = -log [H^{+}]\\2.7 = -log [H^{+}]\\conc. of H^{+} = 1.99 \times 10^{-3}$

Thus, we can conclude that the concentration of hydrogen ions for this solution is $1.99 \times 10^{-3}$ .

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Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and

coldgirl [10]

Answer :

a. The oxidizing agent is: $BH_4^-$

b. The reducing agent is: $ClO_3^-$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The given chemical reaction is,

$BH_4^-(aq)+ClO_3^-(aq)\rightarrow H_2BO_3^-(aq)+Cl^-(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $BH_4^-\rightarrow H_2BO_3^-$

Reduction : $ClO_3^-\rightarrow Cl^-$

Now balance oxygen atom on both side.

Oxidation : $BH_4^-\rightarrow H_2BO_3^-+3H_2O$

Reduction : $ClO_3^-+3H_2O\rightarrow Cl^-$

Now balance hydrogen atom on both side.

Oxidation : $BH_4^-+4OH^-\rightarrow H_2BO_3^-+3H_2O$

Reduction : $ClO_3^-+3H_2O\rightarrow Cl^-+6OH^-$

Now balance the charge.

Oxidation : $BH_4^-+4OH^-\rightarrow H_2BO_3^-+3H_2O+4e^-$

Reduction : $ClO_3^-+3H_2O+6e^-\rightarrow Cl^-+6OH^-$

In order to balance the electrons, we multiply the oxidation reaction by 6 and reduction reaction by 4 and then added both equation, we get the balanced redox reaction.

Oxidation : $6BH_4^-+24OH^-\rightarrow 6H_2BO_3^-+18H_2O+24e^-$

Reduction : $4ClO_3^-+12H_2O+24e^-\rightarrow 4Cl^-+24OH^-$

The balanced chemical equation in acidic medium will be,

$6BH_4^-+4ClO_3^-\rightarrow 6H_2BO_3^-+6H_2O+4Cl^-$

$3BH_4^-+2ClO_3^-\rightarrow 3H_2BO_3^-+3H_2O+2Cl^-$

In the redox reaction, $ClO_3^-$ act as oxidizing agent and $BH_4^-$ act as an reducing agent.

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Two compounds are standing at the same temperature. Compound "A" is evaporating more slowly than compound "B." According to the

frosja888 [35]

Answer:

Option D) Compound B may have a lower molecular weight.

Explanation:

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This can further be seen when gasoline and kerosene are placed under same temperature. The gasoline will evaporate faster than kerosene because the molecular weight of the gasoline is low when compared to that of the kerosene.

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aev [14]

The most concentrated solution is b

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The pOH of a solution of KOH is 11.30. What is the [H*] for this solution?​

The pOH of a solution of KOH is 11.30. What is the [H*] for this solution?