Calculation of ion concentration in high two chemistry Solution

authorname (level nine) is wrong.

The correct answer is as follows:

H2SO4 ionizes out H and inhibits the ionization of H2O. Therefore, in this sulfuric acid, OH comes from the ionization of water, and its concentration is the same as the concentration of H of water ionization, so OH 10 A, H is mainly produced by the ionization of H2SO4, which is much greater than 10 A.

Al2(SO4)3 solution, Al3+ hydrolysis promotes the ionization of water. So in this solution, h is 10 b.

Since the pH of these two solutions is the same, H is also the same, and H is also 10 B in H2SO4 solution. then in H2SO4 solution, KW H·OH 10 a)·(10 b) 10 (a b).

Acids or bases inhibit the ionization of water, and salts (except for strong acids and alkaline salts) promote the ionization of water.

In sulfuric acid solution, the concentration of hydrogen ions ionized by water is 10 -A mol L, the concentration of hydroxide ions ionized by water is 10 -A mol L, and the total hydrogen ion concentration is approximately equal to 10 (a-14) mol L, pH=14-a

Thus, in Al2(SO4)3 solution, the concentration of hydrogen ions ionized from water is 10 (A-14) mol L, pH = 14-A, Al(3+)+3H2O Al(OH)3+3H(+).

1GCA2+, then 1 40=

then no3 - for.

Then the concentration is Zheng Jing's chain manuscript noisy * 1000 = mol l and then ask Sun: Why does no3- be multiplied by 2?

It is the same as finding the number of atoms that I learned earlier.

The concentration of 1mol l mgcl2 mg2+ 1 * 1mol l = 1 cl- concentration = 2 * 1mol l = 2mol l

It is the concentration multiplied by the number of targets below.

Child, this ... Isn't there a formula in the book c=n v, where v is the total volume of the representation, and n is the molar mass of the ion... Analyze the specific situation on a case-by-case basis, find some example questions in the reference book, and don't seek more understanding.

The general idea of calculating the concentration of the amount of ions in the solution of the substance is as follows:

Find the concentration of the solute by multiplying the number of ions by the number of ions in the solute.

For example, if 106GN2CO3 is dissolved in water, find the concentration of Na+ and CO32-?

106Gna2CO3 is 1mol, C(Na2CO3) = 1mol 1L = 1mol L There are two Na+ and one CO32- in a Na2CO3, so C(Na+) = 1mol L 2 = 2mol L C(CO32-) = 1mol L

The amount of concentration of the substance.

What it means: The physical quantity of solute B is expressed as the amount of solute B contained in a unit volume of solution, and the concentration of a substance called solute B.

Unit: mol l or mol*l -1b of the amount of the substance concentration symbol: cb

The commonly used units are mol l and mol m3.

In a solution of a certain concentration of a substance, the relationship between the amount of the substance of solute B (Nb), the volume of the solution (V) and the concentration of the substance of the solute (CB) can be expressed by the following formula:

Formula (calculation of the concept of the quantity concentration of a substance): CB = NB V Quantity concentration of substance (mol L) = Quantity of solute substance (mol) Volume of solution (L).

There is also the amount concentration of the substance = (density of the substance multiplied by 1000 times the mass fraction) (molar mass of the substance multiplied by 1).

How to calculate the concentration of ions.

First, calculate the amount and concentration of the solute.

Look at how many of these ions are contained in a solute molecule.

For example, if 1mol L of H2SO4 is 2*1mol L=2mol L

The concentration of SO42- is 1*1mol L = 1mol L

The pH of the solution is 13, which means that the hydroxide is moles per liter.

Then the remaining Ba(OH)2 is moles per liter.

The remainder ba(OH)2 = mol.

Total BA(OH)2 is molar.

That is, Ba(OH)2 participating in the reaction is molar.

And the resulting barium sulfate is moles, that is, sulfuric acid is moles. The sulfate ion is moles.

Sulfuric acid consumes Ba(OH)2 moles, and the remaining molar Ba(OH)2 is consumed by hydrochloric acid, so hydrochloric acid is mol. Then the chloride ion is the mole.

The hydrogen ion is the sum of sulfuric acid and hydrochloric acid is mole.

So the sulfate ion concentration is moles per liter.

Chloride concentration in moles per liter.

Hydrogen ion concentration in moles per liter.

Analysis: n[ba(oh)2]=

n(baso4)=

ph=13，c(oh-)=

neutralized oh-:n(oh-)=

c(h+)=:1 neutralization).

The SO42- in the original solution all have BaSO4 precipitate:

c(so42-)=

Conservation according to charge.

c(cl-)=

H2SO4=2(H+) SO4) 2- I don't know if it's the total ion concentration, or if you didn't hit it. Let's just calculate it by the total ion concentration. Assuming that the volume of the solution is 1L, the mass of the solution is, the mass of the solute is 1225 4%=49g, then the amount of sulfuric acid is 49g 98g*mol-1=, then H+ has, sulfate has, then the total ion has, and the concentration is.

CaCl2 = Ca2+ +2 Cl-, then the number of chloride ions is twice that of CaCl2 molecules, and in the same solution, the concentration is also doubled, that is, 5mol l 2 = 10mol l.

1、= 1000* = mol/l

Strictly speaking, only a little more than 2 ions can be ionized per molH2SO4, but secondary schools may think that the second stage of sulfuric acid is also completely ionized, so the ion concentration is: actually between).

2 = 10mol/l

The first question -- you're asking about the concentration of the ions -- the total ion concentration is the one upstairs.

The second problem has nothing to do with the 2 moles of the 2 mol L 5 mol l*2 = 10 mol l 2 here 2 is the result of two chloride ions in a molecule.

1GCA2+, then 1 40=

then no3 - for.

then the concentration is mol l

After the calculation of the compound, it is sufficient to use the ionization equation.

According to the number of measurements, for example, 1mol of Al2(SO)3 is dissolved in water to form 2 liters of solution, that cal2(SO)3=1 2=

1Al2(SO)3=2Al3++3SO42- not to mention weak hydrolysis.

In this solution, cal3+=2cal2(SO)3=CSO42-=3cal2(SO)3=

The highest concentration of ions was Na+

The analytic sodium acetate aqueous solution will have such an ionization equation:

CH3COONA = CH3COO-+Na+, because there are very few hydrogen ions and hydroxide ions in the aqueous solution, because the ionization of water is very small and because CH3COO- will be hydrolyzed: CH3COO-+H+ (in water) = CH3COOH, so the acetate ion concentration will decrease, and the unchanged is NA+, so the highest concentration is NA+

H2O = H++OH- (The degree of ionization is so small that it is negligible).

Is it quantity or quality?

Quantity: na more.

Quality: CH3COOH more.