Balance the complete molecular equation. Before writing the ionic equation, the original molecular equation must be balanced. To do this, it is necessary to place the appropriate coefficients in front of the compounds, so that the number of atoms of each element on the left side is equal to their number on the right side of the equation.
- Write the number of atoms of each element on both sides of the equation.
- Add coefficients in front of the elements (except oxygen and hydrogen) so that the number of atoms of each element on the left and right sides of the equation is the same.
- Balance the hydrogen atoms.
- Balance the oxygen atoms.
- Count the number of atoms of each element on both sides of the equation and make sure it is the same.
- For example, after balancing the equation Cr + NiCl 2 --> CrCl 3 + Ni, we get 2Cr + 3NiCl 2 --> 2CrCl 3 + 3Ni.
Determine what state each substance that participates in the reaction is in. This can often be judged by the conditions of the problem. Eat certain rules, which help determine what state an element or connection is in.
Determine which compounds dissociate (separate into cations and anions) in solution. Upon dissociation, a compound breaks down into positive (cation) and negative (anion) components. These components will then enter the ionic equation of the chemical reaction.
Calculate the charge of each dissociated ion. Remember that metals form positively charged cations, and non-metal atoms turn into negative anions. Determine the charges of elements using the periodic table. It is also necessary to balance all charges in neutral compounds.
Rewrite the equation so that all soluble compounds are separated into individual ions. Anything that dissociates or ionizes (such as strong acids) will split into two separate ions. In this case, the substance will remain in a dissolved state ( rr). Check that the equation is balanced.
- Solids, liquids, gases, weak acids and ionic compounds with low solubility will not change their state and will not separate into ions. Leave them as is.
- The molecular compounds will simply disperse into the solution and their state will change to dissolved ( rr). There are three molecular compounds that Not will go into state ( rr), this is CH 4( G) , C 3 H 8 ( G) and C8H18( and) .
- For the reaction under consideration, the complete ionic equation will be written as the following form: 2Cr ( TV) + 3Ni 2+ ( rr) + 6Cl - ( rr) --> 2Cr 3+ ( rr) + 6Cl - ( rr) + 3Ni ( TV) . If chlorine is not part of the compound, it breaks down into individual atoms, so we multiplied the number of Cl ions by 6 on both sides of the equation.
Combine the same ions on the left and right sides of the equation. You can only cross out those ions that are completely identical on both sides of the equation (have the same charges, subscripts, etc.). Rewrite the equation without these ions.
- In our example, both sides of the equation contain 6 Cl - ions, which can be crossed out. Thus, we obtain a short ionic equation: 2Cr ( TV) + 3Ni 2+ ( rr) --> 2Cr 3+ ( rr) + 3Ni ( TV) .
- Check the result. The total charges of the left and right parts ionic equations must be equal.
Instructions
On the left side of the equation, write the substances involved in the chemical reaction. They are called "raw materials". On the right side, respectively, are the formed substances (“reaction products”).
The number of atoms of all elements on the left and right sides of the reaction must be . If necessary, “balance” the quantity by selecting coefficients.
When writing an equation for a chemical reaction, first make sure that it is possible at all. That is, that its occurrence does not contradict the known physical and chemical rules and properties of substances. For example, the reaction:
NaI + AgNO3 = NaNO3 + AgI
It proceeds quickly and completely; during the reaction, an insoluble light yellow precipitate of silver iodide is formed. And the reverse reaction:
AgI + NaNO3 = AgNO3 + NaI - is impossible, although it is written in the correct symbols, and the number of atoms of all elements on the left and right sides is the same.
Write the equation in "full" form, that is, using their molecular formulas. For example, the reaction for the formation of sulfate precipitate:
BaCl2 + Na2SO4 = 2NaCl + BaSO4
Or you can write the same reaction in ionic form:
Ba 2+ + 2Cl- + 2Na+ + SO4 2- = 2Na+ + 2Cl- + BaSO4
In the same way, you can write the equation of another reaction in ionic form. Remember that each molecule of a soluble (dissociating) substance is written in ionic form, identical ions on the left and right sides of the equation are excluded.
A tangent to a curve is a straight line that is adjacent to this curve at given point, that is, it passes through it in such a way that small area around this point you can replace the curve with a tangent segment without much loss of accuracy. If this curve is a graph of a function, then the tangent to it can be constructed using a special equation.
Instructions
Let's say you have a graph of some function. Through two points lying on this, a straight line can be drawn. Such a line intersecting the graph of a given function at two points is called a secant.
If, leaving the first point in place, you gradually move the second point in its direction, then the secant will gradually begin to rotate, tending to some specific position. Eventually, when the two points merge into one, the secant will fit snugly against yours at that single point. Otherwise, the secant will turn into a tangent.
Any inclined (that is, not vertical) line on the coordinate plane is a graph of the equation y = kx + b. The secant passing through the points (x1, y1) and (x2, y2) must therefore satisfy the conditions:
kx1 + b = y1, kx2 + b = y2.
Solving this system of two linear equations, we get: kx2 - kx1 = y2 - y1. Thus k = (y2 - y1)/(x2 - x1).
When the distance between x1 and x2 approaches zero, the differences turn into differentials. Thus, in the equation of the tangent passing through the point (x0, y0), the coefficient k will be equal to ∂y0/∂x0 = f′(x0), that is, the value of the derivative of the function f(x) at the point x0.
To find out the coefficient b, we substitute the already calculated value of k into the equation f′(x0)*x0 + b = f(x0). Solving this equation for b, we get that b = f(x0) - f′(x0)*x0.
As an example, consider the equation of the tangent to the function f(x) = x^2 at the point x0 = 3. The derivative of x^2 is equal to 2x. Therefore, the tangent equation takes the form:
y = 6*(x - 3) + 9 = 6x - 9.
The correctness of this equation is easy
>> Chemistry: Ionic equations
Ionic equations
As you already know from previous chemistry lessons, most chemical reactions occur in solutions. And since all electrolyte solutions include ions, we can say that reactions in electrolyte solutions are reduced to reactions between ions.
These reactions that occur between ions are called ionic reactions. And ionic equations are precisely the equations of these reactions.
As a rule, ionic reaction equations are obtained from molecular equations, but this occurs subject to the following rules:
Firstly, the formulas of weak electrolytes, as well as insoluble and slightly soluble substances, gases, oxides, etc. are not recorded in the form of ions; the exception to this rule is the HSO−4 ion, and then in diluted form.
Secondly, the formulas of strong acids, alkalis, and also water-soluble salts are usually presented in the form of ions. It should also be noted that a formula such as Ca(OH)2 is presented in the form of ions if lime water is used. If lime milk is used, which contains insoluble Ca(OH)2 particles, then the formula in the form of ions is also not written down.
When composing ionic equations, as a rule, the full ionic and abbreviated, that is, brief ionic reaction equations are used. If we consider the ionic equation, which has an abbreviated form, then we do not observe ions in it, that is, they are absent from both parts of the complete ionic equation.
Let's look at examples of how molecular, full and abbreviated ionic equations are written:
Therefore, it should be remembered that the formulas of substances that do not decompose, as well as insoluble and gaseous ones, when drawing up ionic equations are usually written in molecular form.
Also, it should be remembered that if a substance precipitates, a downward arrow (↓) is drawn next to such a formula. Well, in the case when a gaseous substance is released during the reaction, then next to the formula there should be an icon like an upward arrow ().
Let's take a closer look with an example. If we have a solution of sodium sulfate Na2SO4, and we add a solution of barium chloride BaCl2 to it (Fig. 132), we will see that we have formed a white precipitate of barium sulfate BaSO4.
Look closely at the image that shows the interaction between sodium sulfate and barium chloride:
Now let's write the molecular equation for the reaction:
Well, now let's rewrite this equation, where strong electrolytes will be depicted in the form of ions, and reactions that leave the sphere are presented in the form of molecules:
We have written down the complete ionic equation for the reaction.
Now let’s try to remove identical ions from one and the other part of the equality, that is, those ions that do not take part in the reaction 2Na+ and 2Cl, then we will get an abbreviated ionic equation of the reaction, which will look like this:
From this equation we see that the whole essence of this reaction comes down to the interaction of barium ions Ba2+ and sulfate ions
and that as a result, a BaSO4 precipitate is formed, even regardless of which electrolytes contained these ions before the reaction.
How to solve ionic equations
And finally, let's summarize our lesson and determine how to solve ionic equations. You and I already know that all reactions that occur in electrolyte solutions between ions are ionic reactions. These reactions are usually solved or described using ionic equations.
Also, it should be remembered that all those compounds that are volatile, difficult to dissolve or slightly dissociated find a solution in molecular form. Also, we should not forget that in the case when none of the above types of compounds are formed during the interaction of electrolyte solutions, this means that the reactions practically do not occur.
Rules for solving ionic equations
For clear example Let us take the formation of a sparingly soluble compound such as:
Na2SO4 + BaCl2 = BaSO4 + 2NaCl
In ionic form, this expression will look like:
2Na+ +SO42- + Ba2+ + 2Cl- = BaSO4 + 2Na+ + 2Cl-
Since you and I observe that only barium ions and sulfate ions reacted, and the remaining ions did not react and their state remained the same. It follows from this that we can simplify this equation and write it in abbreviated form:
Ba2+ + SO42- = BaSO4
Now let's remember what we should do when solving ionic equations:
First, it is necessary to eliminate the same ions from both sides of the equation;
Secondly, we should not forget that the amount electric charges the equation must be the same, both on its right side and also on its left.
In electrolyte solutions, reactions occur between hydrated ions, which is why they are called ionic reactions. towards them important have the nature and strength of the chemical bond in the reaction products. Typically, exchange in electrolyte solutions results in the formation of a compound with a stronger chemical bond. Thus, when solutions of barium chloride salts BaCl 2 and potassium sulfate K 2 SO 4 interact, the mixture will contain four types of hydrated ions Ba 2 + (H 2 O)n, Cl - (H 2 O)m, K + (H 2 O) p, SO 2 -4 (H 2 O)q, between which the reaction will occur according to the equation:
BaCl 2 +K 2 SO 4 =BaSO 4 +2КCl
Barium sulfate will precipitate in the form of a precipitate, in the crystals of which the chemical bond between the Ba 2+ and SO 2- 4 ions is stronger than the bond with the water molecules hydrating them. The connection between the K+ and Cl - ions only slightly exceeds the sum of their hydration energies, so the collision of these ions will not lead to the formation of a precipitate.
Therefore, we can draw the following conclusion. Exchange reactions occur during the interaction of such ions, the binding energy between which in the reaction product is much greater than the sum of their hydration energies.
Ion exchange reactions are described by ionic equations. Sparingly soluble, volatile and slightly dissociated compounds are written in molecular form. If, during the interaction of electrolyte solutions, none of the indicated types of compounds are formed, this means that practically no reaction occurs.
Formation of sparingly soluble compounds
For example, the interaction between sodium carbonate and barium chloride in the form of a molecular equation will be written as follows:
Na 2 CO 3 + BaCl 2 = BaCO 3 + 2NaCl or in the form:
2Na + +CO 2- 3 +Ba 2+ +2Сl - = BaCO 3 + 2Na + +2Сl -
Only the Ba 2+ and CO -2 ions reacted, the state of the remaining ions did not change, so the short ionic equation will take the form:
CO 2- 3 +Ba 2+ =BaCO 3
Formation of Volatile Substances
The molecular equation for the interaction of calcium carbonate and hydrochloric acid will be written as follows:
CaCO 3 +2HCl=CaCl 2 +H 2 O+CO 2
One of the reaction products - carbon dioxide CO 2 - was released from the reaction sphere in the form of a gas. The expanded ionic equation is:
CaCO 3 +2H + +2Cl - = Ca 2+ +2Cl - +H 2 O+CO 2
The result of the reaction is described by the following short ionic equation:
CaCO 3 +2H + =Ca 2+ +H 2 O+CO 2
Formation of a slightly dissociated compound
An example of such a reaction is any neutralization reaction, resulting in the formation of water, a slightly dissociated compound:
NaOH+HCl=NaCl+H 2 O
Na + +OH-+H + +Cl - = Na + +Cl - +H 2 O
OH-+H+=H 2 O
From the brief ionic equation it follows that the process is expressed in the interaction of H+ and OH- ions.
All three types of reactions proceed irreversibly to completion.
If you merge solutions of, for example, sodium chloride and calcium nitrate, then, as the ionic equation shows, no reaction will occur, since no precipitate, no gas, or low-dissociating compound is formed:
Using the solubility table, we establish that AgNO 3, KCl, KNO 3 are soluble compounds, AgCl is an insoluble substance.
We create an ionic equation for the reaction taking into account the solubility of the compounds:
A brief ionic equation reveals the essence of the chemical transformation taking place. It can be seen that only Ag+ and Cl - ions actually took part in the reaction. The remaining ions remained unchanged.
Example 2. Make up a molecular and ionic equation for the reaction between: a) iron (III) chloride and potassium hydroxide; b) potassium sulfate and zinc iodide.
a) We compose the molecular equation for the reaction between FeCl 3 and KOH:
Using the solubility table, we establish that of the resulting compounds, only iron hydroxide Fe(OH) 3 is insoluble. We compose the ionic equation of the reaction:
The ionic equation shows that the coefficients of 3 in the molecular equation apply equally to ions. This general rule drawing up ionic equations. Let us represent the reaction equation in short ionic form:
This equation shows that only Fe3+ and OH- ions took part in the reaction.
b) Let's create a molecular equation for the second reaction:
K 2 SO 4 + ZnI 2 = 2KI + ZnSO 4
From the solubility table it follows that the starting and resulting compounds are soluble, therefore the reaction is reversible and does not reach completion. Indeed, no precipitate, no gaseous compound, or slightly dissociated compound is formed here. Let's create a complete ionic equation for the reaction:
2K + +SO 2- 4 +Zn 2+ +2I - + 2K + + 2I - +Zn 2+ +SO 2- 4
Example 3. Using the ionic equation: Cu 2+ +S 2- -= CuS, create a molecular equation for the reaction.
The ionic equation shows that on the left side of the equation there must be molecules of compounds containing Cu 2+ and S 2- ions. These substances must be soluble in water.
According to the solubility table, we will select two soluble compounds, which include the Cu 2+ cation and the S 2- anion. Let's create a molecular equation for the reaction between these compounds:
CuSO 4 +Na 2 S CuS+Na 2 SO 4
2.6 Ionic-molecular equations
When any strong acid is neutralized by any strong base, about 57.6 kJ of heat is released for each mole of water formed:
HCl + NaOH = NaCl + H 2 O + 57.53 kJ
HNO 3 + KOH = KNO 3 + H 2 O +57.61 kJ
This suggests that such reactions are reduced to one process. We will obtain the equation for this process if we consider in more detail one of the given reactions, for example, the first. Let's rewrite its equation, writing strong electrolytes in ionic form, since they exist in solution in the form of ions, and weak electrolytes in molecular form, since they are in solution mainly in the form of molecules (water is a very weak electrolyte):
H + + Cl - + Na + + OH - = Na + + Cl - + H 2 O
Considering the resulting equation, we see that during the reaction the Na + and Cl - ions did not undergo changes. Therefore, we will rewrite the equation again, eliminating these ions from both sides of the equation. We get:
H + + OH - = H 2 O
Thus, the reactions of neutralization of any strong acid with any strong base come down to the same process - the formation of water molecules from hydrogen ions and hydroxide ions. It is clear that the thermal effects of these reactions must also be the same.
Strictly speaking, the reaction of the formation of water from ions is reversible, which can be expressed by the equation
H + + OH - ↔ H 2 O
However, as we will see below, water is a very weak electrolyte and dissociates only to a negligible extent. In other words, the equilibrium between water molecules and ions is strongly shifted towards the formation of molecules. Therefore, in practice, the reaction of neutralization of a strong acid with a strong base proceeds to completion.
When mixing a solution of any silver salt with hydrochloric acid or with a solution of any of its salts, a characteristic white cheesy precipitate of silver chloride is always formed:
AgNO 3 + HC1 = AgCl↓ + HNO 3
Ag 2 SO 4 + CuCl 2 = 2AgCl↓ + CuSO 4
Such reactions also come down to one process. In order to obtain its ionic-molecular equation, we rewrite, for example, the equation of the first reaction, writing strong electrolytes, as in the previous example, in ionic form, and the substance in the sediment in molecular form:
Ag + + NO 3 - + H + + C1 - = AgCl↓+ H + + NO 3 -
As can be seen, the H + and NO 3 - ions do not undergo changes during the reaction. Therefore, we exclude them and rewrite the equation again:
Ag + + С1 - = AgCl↓
This is the ion-molecular equation of the process under consideration.
Here it must also be borne in mind that the silver chloride precipitate is in equilibrium with the Ag + and C1 - ions in solution, so that the process expressed by the last equation is reversible:
Ag + + C1 - ↔ AgCl↓
However, due to the low solubility of silver chloride, this equilibrium is very strongly shifted to the right. Therefore, we can assume that the reaction of the formation of AgCl from ions is almost complete.
The formation of an AgCl precipitate will always be observed when there are significant concentrations of Ag + and C1 - ions in the same solution. Therefore, using silver ions, you can detect the presence of C1 - ions in a solution and, conversely, using chloride ions - the presence of silver ions; the C1 - ion can serve as a reagent for the Ag + ion, and the Ag + ion can serve as a reagent for the C1 ion.
In the future, we will widely use the ionic-molecular form of writing equations for reactions involving electrolytes.
To draw up ion-molecular equations, you need to know which salts are soluble in water and which are practically insoluble. general characteristics solubility in water essential salts is given in Table 2.
Ionic-molecular equations help to understand the characteristics of reactions between electrolytes. Let us consider, as an example, several reactions occurring with the participation of weak acids and bases.
Table 2. Solubility of the most important salts in water
As already mentioned, the neutralization of any strong acid by any strong base is accompanied by the same thermal effect, since it comes down to the same process - the formation of water molecules from hydrogen ions and hydroxide ions. However, when neutralizing a strong acid with a weak base, or a weak acid with a strong or weak base, the thermal effects are different. Let's write ion-molecular equations for such reactions.
Neutralization of a weak acid (acetic acid) with a strong base (sodium hydroxide):
CH 3 COOH + NaOH = CH 3 COONa + H 2 O
Here the strong electrolytes are sodium hydroxide and the resulting salt, and the weak ones are acid and water:
CH 3 COOH + Na + + OH - = CH 3 COO - + Na + + H 2 O
As can be seen, only sodium ions do not undergo changes during the reaction. Therefore, the ion-molecular equation has the form:
CH 3 COOH + OH - = CH 3 COO - + H 2 O
Neutralization of a strong acid (nitrogen) with a weak base (ammonium hydroxide):
HNO 3 + NH 4 OH = NH 4 NO 3 + H 2 O
Here we must write the acid and the resulting salt in the form of ions, and ammonium hydroxide and water in the form of molecules:
H + + NO 3 - + NH 4 OH = NH 4 - + NH 3 - + H 2 O
NO 3 - ions do not undergo changes. Omitting them, we obtain the ionic-molecular equation:
H + + NH 4 OH= NH 4 + + H 2 O
Neutralization of a weak acid (acetic acid) with a weak base (ammonium hydroxide):
CH 3 COOH + NH 4 OH = CH 3 COONH 4 + H 2 O
In this reaction, all substances, except the salt formed, are weak electrolytes. Therefore, the ion-molecular form of the equation looks like:
CH 3 COOH + NH 4 OH = CH 3 COO - + NH 4 + + H 2 O
Comparing the obtained ion-molecular equations with each other, we see that they are all different. Therefore, it is clear that the heats of the reactions considered are also different.
Reactions of neutralization of strong acids with strong bases, during which hydrogen ions and hydroxide ions combine to form a water molecule, proceed almost to completion. Neutralization reactions in which at least one of the starting substances is a weak electrolyte and in which molecules of weakly dissociating substances are present not only on the right side, but also on the left side ion-molecular equation, do not proceed completely. They reach a state of equilibrium in which the salt coexists with the acid and base from which it was formed. Therefore, it is more correct to write the equations of such reactions as reversible reactions:
CH 3 COOH + OH - ↔ CH 3 COO - + H 2 O
H + + NH 4 OH↔ NH 4 + + H 2 O
CH 3 COOH + NH 4 OH ↔ CH 3 COO - + NH 4 + + H 2 O
With other solvents, the considered patterns remain the same, but there are also deviations from them, for example, a minimum (anomalous electrical conductivity) is often observed on the λ-c curves. 2. Ion mobility Let us relate the electrical conductivity of the electrolyte to the speed of movement of its ions in electric field. To calculate electrical conductivity, it is enough to count the number of ions...
When studying the synthesis of new materials and processes of ion transport in them. IN pure form Such patterns are most clearly visible in the study of single-crystal solid electrolytes. At the same time, when using solid electrolytes as working media for functional elements, it is necessary to take into account that materials of a given type and shape are needed, for example in the form of dense ceramics...
17-25 kg/t aluminum, which is ~ 10-15 kg/t higher compared to the results for sandy alumina. Alumina used for aluminum production must contain a minimum amount of iron, silicon, heavy metals with a lower release potential at the cathode than aluminum, because they are easily reduced and converted into cathode aluminum. It is also undesirable to be present in...