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## What volume of 10^(-3)M HCl should be added to 1000 ml. of 10^(-3)M NaOH, so that pH is reduced by 1 unit? Thank you:)

Michael
Featured 4 months ago

818 ml.

#### Explanation:

First we find the pH of solution:

$\textsf{\left[O {H}^{-}\right] = {10}^{- 3} \textcolor{w h i t e}{x} M}$

The ionic product of water gives us:

$\textsf{\left[{H}^{+}\right] \left[O {H}^{-}\right] = {10}^{- 14}}$

$\therefore$$\textsf{\left[{H}^{+}\right] = {10}^{- 14} / \left[\left[O {H}^{-}\right]\right] = {10}^{- 14} / {10}^{- 3} = {10}^{- 11} \textcolor{w h i t e}{x} M}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left[{10}^{- 11}\right] = 11}$

This means we have to create a solution where the pH = 10.

Adding $\textsf{{H}^{+}}$ ions will have 2 effects:

1. $\textsf{O {H}^{-}}$ ions will be removed due to $\textsf{{H}^{+} + O {H}^{-} \rightarrow {H}_{2} O}$. This will lower their concentration.

2. The total volume of the solution will increase. This will also reduce $\textsf{\left[O {H}^{-}\right]}$.

We need to find the volume of acid V to add.

Since the target pH = 10 this means that $\textsf{\left[{H}^{+}\right] = {10}^{- 10} \textcolor{w h i t e}{x} M}$

$\therefore$$\textsf{\left[O {H}^{-}\right] = {10}^{- 14} / {10}^{- 10} = {10}^{- 4} \textcolor{w h i t e}{x} M}$

The total moles of $\textsf{O {H}^{-}}$ in the original solution is given by:

$\textsf{{n}_{O {H}^{-}} \text{initial} = c \times v = {10}^{- 3} \times 1 = {10}^{- 3} = 0.001}$ mol

The number of moles of $\textsf{{H}^{+}}$ added is given by:

$\textsf{{n}_{{H}^{+}} = c \times V = 0.001 \times V}$ mol

Since they react in the ratio 1:1 the number of moles of $\textsf{O {H}^{-}}$ consumed = $\textsf{0.001 \times V}$ mol.

So the number of moles of $\textsf{O {H}^{-}}$ remaining is given by

$\textsf{{n}_{O {H}^{-}} = 0.001 - 0.001 V}$

In litres the total new volume = $\textsf{\left(1 + V\right)}$

This means the required concentration of $\textsf{{10}^{- 4} \textcolor{w h i t e}{x} M}$ for $\textsf{O {H}^{-}}$ can be written:

$\textsf{\left[O {H}^{-}\right] = \frac{n}{v} = \frac{\left(0.001 - 0.001 V\right)}{\left(1 + V\right)} = {10}^{- 4} \textcolor{w h i t e}{x} M}$

$\therefore$$\textsf{\left(0.001 - 0.001 V\right) = {10}^{- 4} \left(1 + V\right)}$

$\textsf{0.001 \left(1 - V\right) = {10}^{- 4} \left(1 + V\right)}$

$\textsf{\frac{\left(1 - V\right)}{\left(1 + V\right)} = {10}^{- 4} / \left({10}^{- 3}\right) = 0.1}$

$\textsf{\left(1 - V\right) = 0.1 \left(1 + V\right)}$

$\textsf{1 - V = 0.1 + 0.1 V}$

$\textsf{1.1 V = 0.9}$

$\textsf{V = \frac{0.9}{1.1} = 0.818 \textcolor{w h i t e}{x} L}$

$\textsf{V = 818 \textcolor{w h i t e}{x} m l}$

Check by iteration:

$\textsf{\left[O {H}^{-}\right] = \frac{0.001 - 0.001 \times 0.818}{1 + 0.818}}$

$\textsf{\left[O {H}^{-}\right] = \frac{\cancel{1.818} \times {10}^{- 4}}{\cancel{1.818}} = {10}^{- 4}}$

$\therefore$$\textsf{\left[{H}^{+}\right] = {10}^{- 14} / \left(\left[O {H}^{-}\right]\right) = {10}^{- 14} / \left({10}^{- 4}\right) = {10}^{- 10} \textcolor{w h i t e}{x} M}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left[{10}^{-} 10\right] = 10}$

So that's all good.

## Please answer this question: Pentane is a volatile water-insoluble compound with a very low surface tension. Explain these properties!? Thank you!

anor277
Featured 3 months ago

Consider the $\text{degree of intermolecular interaction}$ between pentane molecules.....

#### Explanation:

And we distinguish between $\text{intermolecular force}$, the force between molecules, and $\text{intramolecular force}$, the force between atoms WITHIN molecules.

For pentane, for any of its three structural isomers (which are?), the $\text{intramolecular force}$ is STRONG. That is $C - H$ and $C - C$ bonds are thermodynamically strong; they are not as strong as $C = O$, and $O - H$ bonds, which is why we use hydrocarbons as a source of energy.

On the other hand, the $\text{intermolecular force}$, relies on dispersion force between the pentane strands....longer strands generate greater intermolecular force, and this is nicely indicated by the normal boiling points, $\text{n-pentane} , 36.1$ ""^@C; versus $\text{isopentane} , 27.7$ ""^@C; $\text{neopentane} , 9.5$ ""^@C. The more branching the LESS the degree of intermolecular interaction.

And as regards solubility in water, there is little for water to work on as a solvent. The pentanes are molecular, and certainly they are non-polar, and hence addition of $\text{n-pentane}$ to water gives rise to two immiscible layers, with the LESS dense pentane floating on top of the funnel. The reduced surface tension of pentane, can also be attributed to the lack of intermolecular interaction.

And contrast this with the solubility of ethanol and methanol in water. What has an alcohol got that an alkane ain't got?

## Why is the first ionization energy of a nonmetal much higher than that of an alkali metal?

Meave60
Featured 3 months ago

Refer to the explanation.

#### Explanation:

The first ionization energy of an element is the energy needed to remove the outermost (valence) electron from a neutral atom in the gaseous state to form a cation.

${X}_{\text{(g)" + "energy}}$$\rightarrow$$\text{X"_"(g)"^(+) + e^(-)}$

The first ionization energy of an element has an inverse relationship to its atomic radius.

Across a period from left to right on the periodic table, first ionization energy increases and atomic radius decreases. Down a group, first ionization energy decreases and atomic radius increases.

The atoms of the alkali metals have larger radii than those of nonmetals. Because of this, the attraction of the positively charged atomic nucleus for the valence electron is much less than that of a nonmetal.

Therefore, the first ionization energy of an alkali metal is much less than that of a nonmetal. Because of this, alkali metals lose their single valence electron which produces a cation with a charge of ${1}^{+}$. An example is the neutral sodium (Na) atom, in period 3, which has an atomic radius of $\text{190 picometers}$ $\left(\text{pm}\right)$, and a first ionization energy of $\text{496 kilojoules/mol}$ $\left(\text{kJ/mol}\right)$.

The atoms of nonmetals have smaller radii, therefore the positively charged nucleus has a greater attraction for the valence electrons.

So the first ionization energy for a nonmetal is much greater than that of an alkali metal, so nonmetals do not lose electrons in an ionic bond, but instead gain one or more electrons and form anions. An example is a neutral chlorine (Cl) atom, also in period 3, which has an atomic radius of $\text{79 pm}$, and a first ionization energy of $\text{1251 kJ/mol}$. Compare these values with a neutral Na atom previously mentioned.

Refer to the diagram below.

The first ionization energy increases across a period. $\text{H}$ (not labeled) to $\text{He}$ is period 1, $\text{Li}$ to $\text{Ne}$ is period 2, $\text{Na}$ to $\text{Ar}$ is period 3, and so on.

## What are low and high spin complexes?

Truong-Son N.
Featured 2 months ago

It just categorizes, qualitatively, how the metal $d$ orbitals are filled in crystal field theory after they are split by what the theory proposes are the ligand-induced electron repulsions.

The usual Hund's rule and Aufbau Principle apply.

Examples of low-spin ${d}^{6}$ complexes are ["Cr"("CN")_6]^(3-) and "Cr"("CO")_6, and examples of high-spin ${d}^{6}$ complexes are ${\left[{\text{CrCl}}_{6}\right]}^{3 -}$ and "Cr"("H"_2"O")_6.

I assume you know the basic facets of crystal field theory:

1. Ligands come in, and their important orbitals interact with the metal $d$ orbitals.
2. Electrons repel electrons to destabilize certain metal $d$ orbitals. In an octahedral field, these are known as the ${e}_{g}^{\text{*}}$ orbitals.
3. Electrons are attracted to the electropositive metal center to stabilize certain metal $d$ orbitals. In an octahedral field, these are known as the ${t}_{2 g}$ orbitals.

The crystal field splitting energy is called ${\Delta}_{o}$ in an octahedral field for simplicity, and the resultant $d$ orbital splitting is:

uarrE" "color(white)({(" "" "color(black)(ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "e_g^"*")),(color(black)(Delta_o)),(" "color(black)(ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "t_(2g))):})

Orbitals close in energy simultaneously fill more easily and vice versa. And so, depending on the magnitude of ${\Delta}_{o}$, there are two cases. Take a ${d}^{6}$ configuration as an example...

• When ${\Delta}_{o}$ is large, the complex is likely low-spin:

uarrE" "color(white)({(" "" "color(black)(ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "e_g^"*")),(),(),(),(),(color(black)(Delta_o)),(),(),(),(),(" "color(black)(ul(uarr darr)" "ul(uarr darr)" "ul(uarr darr)" "t_(2g))):})

• When ${\Delta}_{o}$ is small, the complex is likely high-spin:

uarrE" "color(white)({(" "" "color(black)(ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "e_g^"*")),(),(color(black)(Delta_o)),(),(" "color(black)(ul(uarr darr)" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "t_(2g))):})

Of course, I am exaggerating the energy scale, but hopefully that brings the point across.

• High spin complexes half-fill the lower energy $\boldsymbol{d}$ orbitals first, and THEN move up to the higher energy $d$ orbitals to half-fill those next, before pairing starts occurring, BECAUSE those orbitals are so similar in energy to the lower energy orbitals.
• Low spin complexes fill the lower energy orbitals completely first, before moving on to the higher energy orbitals, BECAUSE those orbitals are so much higher in energy.

## Hello, I know what amphoteric oxides are, but what factor determines whether an oxide is amphoteric?

Ernest Z.
Featured 1 month ago

Warning! Long Answer. Here's my explanation.

#### Explanation:

Basic oxides

Metallic character increases from right to left and from top to bottom in the Periodic Table.

The most metallic elements form the most basic oxides.

Even if the oxides are insoluble in water, we still call them basic oxides because they react with acids.

$\text{MgO(s) + 2HCl(aq) → MgCl"_2"(aq)" + "H"_2"O"(l)}$

Acidic oxides

Nonmetallic character increases from left to right and from bottom to top in the Periodic Table.

The most nonmetallic elements form the most acidic oxides.

They react with water to form oxoacids. For example,

$\text{SO"_2"(aq)" + "H"_2"O(l)" → "H"_2"SO"_3"(aq)}$

Even if an oxide is insoluble on water, we still class it as acidic if it reacts with bases to form salts. For example,

$\text{TeO"_2"(s)" + "2NaOH(aq)" → "Na"_2"TeO"_3"(aq)" + "H"_2"O(l)}$

Amphoteric Oxides

Some oxides react with both acids and bases, that is, they are amphoteric.

For example, ${\text{Al"_2"O}}_{3}$ is an amphoteric oxide.

$\text{Al"_2"O"_3"(s)" + "6H"_3"O"^"+""(aq)" + "3H"_2"O(l)" → 2"Al"("OH"_2)_6^"3+""(aq)}$
$\text{Al"_2"O"_3"(s)" + 2"OH"^"-""(aq)" + "3H"_2"O(l)" → 2"Al"("OH")_4^"-""(aq)}$

The lighter elements of Groups 2 and 13, some of the $\text{d}$-block elements, and the heavier elements of Groups 14 and 15 have amphoteric oxides.

The most basic oxides are at the lower left of the Periodic Table and the most acidic oxides are at the upper right, so it is not surprising that the borderline between acidic and basic oxides occurs along a diagonal.

Amphoterism and oxidation states

Amphoterism depends on the oxidation state of the oxide.

There is no simple way to predict which elements will be amphoteric.

The amphoteric character of an oxide probably reflects the ability of the metal to polarize the surrounding oxide ions, that is, to introduce significant covalent character into the $\text{M-O}$ bond.

This ability increases with oxidation state as the positive character of the central atom increases.

However, in Group 15, only the oxides with lower oxidation states are amphoteric.

The oxides with higher oxidation states are too acidic to be amphoteric.

## "MnO"_4^(2-) -> "MnO"_4^(-) + "MnO"_2. How in the world do I begin to half reduce this? There are two elements being oxidized. So there aren't two ways to oxidize it. Or is there?

Stefan V.
Featured 3 weeks ago

Here's what's going on here.

#### Explanation:

You're actually dealing with a disproportionation reaction here. In a disproportionation reaction, the same element undergoes both oxidation and reduction.

In this case, manganese(VI) is reduced to manganese(IV) and oxidized to manganese(VII).

stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + stackrel(color(blue)(+4))("Mn")"O"_ (2(s))

The reduction half-reaction looks like this

stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s))

Here each atom of manganese takes in $2$ electrons, which is why the oxidation number of manganese goes from $\textcolor{b l u e}{+ 6}$ on the reactants' side to $\textcolor{b l u e}{+ 4}$ on the products' side.

To balance the atoms of oxygen, use the fact that this reaction takes place in an acidic medium and add water molecules to the side that needs oxygen and protons, ${\text{H}}^{+}$, to the side that needs hydrogen.

$4 {\text{H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O}}_{\left(l\right)}$

Notice that the half-reaction is balanced in terms of charge because you have

$4 \times \left(1 +\right) + \left(2 -\right) + 2 \times \left(1 -\right) = 0 + 0$

$\textcolor{w h i t e}{a}$
The oxidation half-reaction looks like this

stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + "e"^(-)

This time, each atom of manganese loses $1$ electron, which is why the oxidation number of manganese goes from $\textcolor{b l u e}{+ 6}$ on the reactants' side to $\textcolor{b l u e}{+ 7}$ on the products' side.

The atoms of oxygen are already balanced, so you don't need to use water molecules and protons. Once again, the half-reaction is balanced in terms of charge because you have

$\left(2 -\right) = \left(1 -\right) + \left(1 -\right)$

So, you know that the balanced half-reactions look like this

$\left\{\begin{matrix}4 {\text{H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l)) \\ color(white)(aaaaaaaaaaaaaa)stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + "e}}^{-}\end{matrix}\right.$

In every redox reaction, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction, so multiply the oxidation half-reaction by $2$ to get

$\left\{\begin{matrix}4 {\text{H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l)) \\ color(white)(aaaaaaaaaaaaa)2stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> 2stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + 2"e}}^{-}\end{matrix}\right.$

Add the two half-reactions to find the balanced chemical equation that describes this disproportionation reaction.

$\left\{\begin{matrix}4 {\text{H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l)) \\ color(white)(aaaaaaaaaaaaa)2stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> 2stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + 2"e}}^{-}\end{matrix}\right.$
$\frac{\textcolor{w h i t e}{a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}$
$4 {\text{H"_ ((aq))^(+) + ["MnO"_ (4(aq))^(2-) + 2"MnO"_ (4(aq))^(2-)] + color(red)(cancel(color(black)(2"e"^(-)))) -> 2"MnO"_ (4(aq))^(-) + "MnO"_ (2(s)) + color(red)(cancel(color(black)(2"e"^(-)))) + 2"H"_ 2"O}}_{\left(l\right)}$

You will end up with

$4 {\text{H"_ ((aq))^(+) + 3"MnO"_ (4(aq))^(2-) -> 2"MnO"_ (4(aq))^(-) + "MnO"_ (2(s)) + 2"H"_ 2"O}}_{\left(l\right)}$

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