Chemical ion concentration question, once to the chemical ion concentration question, please help wi

1) 1 1 (2) greater than or equal to.

It is greater than according to the charge balance, c(h+)+c(na+)=c(oh-)+c(a-), and since c(oh-) c(h+), c(na+) c(a-).

b c analysis: (1) h2SO4+2NaOH====Na2SO4+2H2O

mol·l-1

Solution v1=v2

v1∶v2=1∶1。

2) The solute in the obtained solution is only Naa, if it is a strong alkali and strong salt, the solution pH = 7;If it is a strong alkali and weak salt, the hydrolysis is alkaline, and the solution pH is 7. Q, from pH 7, we can know c(oh-) c(ch+), and then from the conservation of charge we know c(na+)+c(h+)=c(oh-)+c(a-), then c(na+) c(a-). Q, C(H+)=C(OH-)=1 10-7 in neutral solution at room temperature

mol·l-1, then c(h+)+c(oh-)=2 10-7

mol·l-1,a is correct. When HA is a weak acid, if V1=V2, the HA is excessive after the reaction, and the resulting solution is acidic, pH 7, B, C are incorrect. HA may be a strong acid or a weak acid, c(ha) c(h+) = 10-pH

mol·l-1=10-3

mol·l-1, while c(naoh) = c(oh-) = 10ph-14

mol·l-1=10-3

mol·L-1, if the mixture is alkaline, the NaOH solution must be excessive, then V1 must be less than V2, and the D term is correct.

V1:V2=1:1, pH=7, greater than, OH value greater than 7 indicates that the solution is alkaline, while sodium ion hydrolysis is alkaline, and weak alkaline after neutralization with weak acid groups. c

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.

When B and pH are both 3, due to the different ionization constants, the volume is V0, so the concentration of formic acid is smaller, so when both are neutralized with NaOH, the formic acid consumes less NaOH and N (Na+) less. I guess so.

Note that the "concentration of hydrogen ions ionized by water" and "hydrogen ion concentration ionized by water" in the stem do not represent the concentration of hydrogen ions in the solution.

The addition of hydrochloric acid is divided into three processes:

1. The amount of hydrochloric acid added is not enough to complete the reaction: at this time, hydrochloric acid is added to react with ammonia, which reduces the "inhibition" of hydrogen ions by hydroxide in the ammonia aqueous solution, so hydrochloric acid is added before point C, "It is an ammonium chloride solution, which is acidic after hydrolysis of strong acid and weak alkali salt, and promotes the electrolysis of hydrogen ions by water."

3. When excessive hydrochloric acid is added, the hydrogen ions in the solution will increase, and the concentration of hydrogen ions in the acidic solution will inhibit the water from continuing to ionize hydrogen ions.

pH refers to the concentration of hydrogen ions in a solution, not "the concentration of hydrogen ions ionized by water", so it is not without two neutrals.

Hope, thank you!

At room temperature:

c[h+] c [oh-] = 10^(-14) mol^2 / l^2

pH Calculation Method:

ph= - log 10 (c(h+))

mol L of HCl solution, whose pH = - log 10 (10 -1) = - 1) = 1

So:1mol L of HCl solution, C(H+) = mol L

So c(oh-) = 10 (-13) mol l

OH- can only be ionized by water, and when water ionizes OH-, it will also ionize H+ by the same amount

So, [H+]=c(oh-)= 10 (-13) mol l

mol L of Ba(OH)2 solution, C(OH-) = mol L

So c(h+) = 10 (-13) mol l

H+ can only be ionized by water, and when water ionizes H+, it will also ionize OH- in equal amounts

So, the [oh-] = c(h+) = 10 (-13) mol l ionized by water

Even if the solution of pure Naha is acidic, in a mixed solution with a concentration of H2A, it is not a superposition of ionization of the two, but inhibits each other. In fact, in the mixed solution of their equal concentration [to form a buffer solution], the ionization of Ha- can also be ignored [because the ionization constants of the adjacent levels of multiple weak acids are usually more than 10,000 times different], and the main equilibrium is still the primary ionization of H2A, due to the presence of a large amount of Ha-, it will be strongly inhibited, so the concentration of hydrogen ion finch fiber in A is higher than that in C.

The concentration of H+ in a is approximately between and is not completely ionized.

The concentration of H+ in B is less than that, and the reason for Peichun is the same as that of A

The HC-level HCl provides the H+ concentration and also inhibits the ionization of NaHA, so the overall H+ concentration is less than that.

Because the neutralization reaction takes place in d, H+ is almost non-existent, so it is minimal.

2.The explanation is the same as (1), because of the neutralization reaction, almost A2- is generated, and the A2A molecule is basically H2A.

A: Depending on the charge balance, this is the obvious result.

D: According to the mass balance, H2S is common at the beginning, and H2S, Hs-, and S2- are obtained after the reaction, and the total amount of these three should also be. So the sum of these three concentrations is 1mol l

C: According to D, C(Na+)=, the sum of the last three is 1mol L, and it is obvious that B: the reaction eventually forms HS- and S2- buffer solutions, which should be alkaline, so C(H+)>C(OH-) are wrong.

A: According to the charge balance, D: According to the mass balance, H2S is common at the beginning, and H2S, Hs-, and S2- are obtained after the reaction, and the total amount of these three should also be. So the sum of these three concentrations is 1mol l

C: According to D,C(Na+)=, the sum of the last three is 1mol L,B: The reaction finally forms HS- and S2- buffer solutions, which should be alkaline, so C(H+)>C(OH-) are wrong.

The concentration of NH4+ in the stem is:

1 in A, 1 in B, and C in C.

D is the reason why BC is chosen

This question won't be a single choice, it's been a long time, and it's a bit vague.

1*2 2 in Question A

B in C. D in the selection of BC Note: do not care about the volume, only look at the concentration.

Hehe, look at the number of nitrogen elements can be solved!