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2 years ago

What would the [OH-] of a solution that has a pOH of 2.7 be?

Chemistry

1 answer:

Brilliant_brown [7]2 years ago

6 0

Answer:

The [OH⁻] of a solution that has a pOH of 2.7 will be 2*10⁻³

Explanation:

The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution.

pOH indicates the concentration of hydroxyl ions [OH-] present in a solution. In this way, pOH is defined as the negative logarithm of the activity of hydroxide ions, that is, the concentration of OH- ions:

pOH= -log [OH⁻]

In this case, pOH has a value of 2.7. Replacing:

2.7= -log [OH⁻]

and solving:

[OH⁻]=10⁻² ⁷

you get:

[OH⁻]≅ 2*10⁻³

The [OH⁻] of a solution that has a pOH of 2.7 will be 2*10⁻³

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2 years ago

If kc = 7.04 × 10-2 for the reaction: 2 hbr(g) ⇌h2(g) + br2(g), what is the value of kc for the reaction: 1/2 h2(g) + 1/2 br2 ⇌h

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At the first reaction when 2HBr(g) ⇄ H2(g) + Br2(g)
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Its Kc value will = [HBr] / [H2]^1/2*[Br2]^1/2
we will make the first formula of Kc upside down:
1/7.04X10^-2 = [HBr]^2/[H2][Br2]
and by taking the square root:
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3 years ago

Write balance complete molecular equation, ionic equation, and net ionic equations for the reactions that occur when each of the

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Answer: The complete molecular, ionic, and net ionic equations are given below. The spectator ions are sodium and nitrate ions.

Explanation:

The ionic equation is defined as the equation in which all the substances that are strong electrolytes present in an aqueous state and are represented in the form of ions.

The net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

The balanced molecular equation for the reaction of lead (II) nitrate and sodium sulfide follows:

$Pb(NO_3)_2(aq)+Na_2S(aq)\rightarrow PbS(s)+2NaNO_3(s)$

The ionic equation follows:

$Pb^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+S^{2-}(aq)\rightarrow PbS(s)+2Na^+(aq)+2NO_3^-(aq)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Pb^{2+}(aq)+S^{2-}(aq)\rightarrow PbS(s)$

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