I am a senior in high school, and I have questions about chemical pH calculations!

I'm three. You can actually know about the conservation of protons on the encyclopedia, the above is more detailed, and I will tell you my method.

The substances in the NaHCO3 solution are H+ OH- Na+ HCO3- carbonate and carbonic acid are present but in smaller amounts.

The equation for the conservation of protons is H+ + Carbonic Acid = Oh- + Carbonate.

Because water electrolysis H2O=OH- +H+ HCO3-=H+ + carbonate Bicarbonate hydrolysis produces carbonic acid and hydroxide.

Proton conservation is actually an equation written from hydrogen ions and hydroxide ions, and the equation for the conservation of protons may not be the same depending on personal habits, but they are all the same.

.3 Yes is the value of the approximate.

7 = 14 is the negative logarithm of the water ion product constant.

Try two specific numbers.

Just use the basic formula.

For the first and second questions, I'll give you a hint, note that the stem says that it is an equal volume mixture, lg2; As for the third question, I don't remember that there was this knowledge point in high school, there is the concept of proton balance in chemistry in college, let's say NaHCO3 solution, you can choose the most primitive acid-base components HCO3 and H2O as the proton reference level, the product after obtaining protons is H+, H2CO3, and the product after losing protons is OH-, CO3 2-, so the proton condition formula is: [H+]+CO2]=[OH-]+CO3 2-].

The above two are empirical formulas, and only apply if the pH difference is greater than or equal to 2.

It is enough to understand these things as a student, and there is no need to grasp them. I haven't seen any of these formulas examined in the exam questions.

False: C(H+)=10 -A, then C(OH-)=10 A-14 N(OH-)=X10 A-14, obtained from 10=Y and A+B=13: N(OH-)=Y10 -B-2 and C(H+)=|0 -b then n(h+) = y10 -b can be neutralized after the partial acid.

The concentration of NaOH with pH=A is 10 (-14+A)mol L, and the concentration of HCl with pH=B is 10-B mol L;

So the total NAOH is 10 (-14+A)*x;

HCL is 10 -b *y=10 -b *10x=10 (-b+1)*x=10 (a-12)*x;

then the total H+ after the reaction is 10 (a-12)*x-10 (-14+a)*x; The solution is acidic;

The concentration of H+ is 11X;

Then you just need to take the logarithm to know the pH value. I don't have a calculator around, so you can press it yourself.

The idea is the same as above, theoretically it should be correct, it has been too long since I graduated, and it is almost ...... to forget

Method 1: Conventional derivation.

pH = A, NaOh, i.e., C(H+) = 10 -A, C(OH-) = 10 - (14-A) = 10 (A-14) mol L

x ml, i.e., n(Naoh) = x*10 -3*10 (A-14) = X*10 (A-17) mol

pH = B, HCl, i.e., C(H+) = 10 -B mol L

Y ml, i.e., N(HCl)=Y*10 -3*10 -B=Y*10 -(B+3) mol

Substitute y=10x, b=13-a.

It is obtained, n(hcl)=10x*10 -(13-a+3)=x*10 (a-15), compared with n(naoh), hcl is more, so, not ph = 7, but, ph < 7

Shouldn't it be, x=10y

In this way, the substitution is, y=x 10, and we get, n(hcl)=x*10 -1*10 -(b+3)=x*10 (a-17)=n(naoh).

It is just neutralized, so, ph=7

Method 2: Here happens to be the strong acid HCl and the strong base NaOH, both of which are ionized in the outer ring, so there is a law, the pH of the acid = B, and the pH of the base = A

When, a+b=14, the equal volume is mixed, which is just neutralized.

If, a+b < 14, the concentration of the acid is greater, then the volume of the base needed to neutralize is just more.

You can deduce this law yourself, and after mastering it, it will be faster to judge such a topic.

No, it should be 10x=y or x=10y

First of all, you need to know that pH is log(h+) in solution, and neutral is pH=7

Because NaOH HCl is a strong electrolyte, it is easier to investigate.

First, calculate the amount of H+ and OH- separately, and then use the difference subtraction method to see which amount is more.

The relationship between pH and concentration is involved, just remember that [h+] = 10 (-ph); oh-] = 10 (ph-14). [h+]*oh-] = 10 ^(14)。

Knowing the above method, you can calculate. This question is calculated as follows:

The molar concentration of NaOH is 10 (-14+A) for pH=A and 10 (-B) for PH=B.

So the quantity relation of the substances of H+ after mixing is.

delta h+ = - x*[10^(a-14)] y*[10^(-b)]

x * 10^(-b-1) +10x * 10^(-b)]

x * x* 。

The concentration of 0 [h+] is equal to the amount of its substance divided by the volume of solution, remember that the volume is x+y milliliters at this point.

Because of the excess acid, the pH of the solution is < 7.

Hello:(1).pH = 12 in NaOH solution, c(OH-) =; In NaOH solution with pH 11, C(OH-)=, according to the dilution formula:

The solution obtained v = 1000ml, that is, the volume of water added = 1000-100 = 900ml.

Solution v=1000ml

3) Set HCL to be added

Solution v=so should be added.

2.(1) At a certain temperature. There is aHydrochloric acid bSulfuric acid cAcetic acid.

The amount and concentration of the three acids are the same. c(h+) in descending order.

Let their concentrations be C, HCl hydrochloride is a strong acid of one element, completely ionized, and the concentration of hydrogen ions is equal to C; H2SO4 sulfate is a dibasic strong acid, and the concentration of hydrogen ions ionized is 2C; Acetic acid is a single weak acid, a weak electrolyte, and the concentration of hydrogen ions ionized is less than C, so the concentration of hydrogen ions of the three is B>A>C

Three acids of the same volume and concentration of the same substance. The ability to neutralize NAOH is from large to small.

The volume and concentration of the three acids are the same, that is, the amount of their substances is the same, and if the amount of the three substances is amol, then HCL--Naoh, AmolnaOH is required

H2SO4 --2NaOH, 2AmolNaOH is required

HAC --- NaOH and requires Amol sodium hydroxide, so it is b>a=c

2) Suppose there is a pH 1 solution of hydrochloric acid, sulfuric acid and acetic acid.

The hydrochloric acid concentration is hydrogen, completely ionized).

The concentration of sulfuric acid is less than that of hydrogen, the first is completely ionized, and the second is nearly completely ionized).

The concentration of acetic acid is much greater than that of weak acid, and there is an ionization equilibrium) is about 1mol l.

The above approximate concentration can be calculated, but the temperature should be considered, and the general temperature is 20-25 degrees.

Therefore, when 1. C(H+) is the same, the quantity and concentration of substances are in order from large to small.

First compare hydrochloric acid and acetic acid, hydrochloric acid is a strong acid, acetic acid is a weak acid, to ionize to obtain the same concentration of hydrogen ions, because acetic acid is not completely ionized, so the concentration of acetic acid is greater than hydrochloric acid, that is, c>a

Now compare hydrochloric acid and sulfuric acid, both are strong acids, to get the same concentration of hydrogen ions, the concentration of sulfuric acid is only half of hydrochloric acid, because sulfuric acid is a dibasic acid, and hydrochloric acid is a monobasic acid, that is, ba>b

2. The c(h+) of the three is the same and the volume is the same. Put in a sufficient amount of zn, and produce gas in the order from large to small under the same condition.

When a sufficient amount of zinc is put in, the volume of the gas produced (that is, hydrogen) is only determined by the amount of hydrogen ion substance, hydrochloric acid and sulfuric acid are both strong acids, which are ionized, and the hydrogen ion concentration and the volume of the solution are the same, so the amount of hydrogen ion in the two is the same, and the volume of hydrogen produced must be the same. Acetic acid is a weak acid, and some hydrogen ions are not ionized, as the reaction progresses, hydrogen ions continue to ionize, and the total amount of hydrogen ions that can be obtained in the end exceeds that of hydrochloric acid and sulfuric acid, so the gas produced is the most. To sum up, this question:

c>a=b

The concentration of hydrogen ions ionized by HCl is:

The concentration of hydrogen ions ionized by water is x

Then there is (x+.)

Solve the equation to obtain: x=

The total concentration of hydrogen ions is:

ph=-lg(

Because it is a strong acid, the hydrogen ion is 5 10 -8 mol l, pH=-log[h+]=, but the acid pH is up to 7, so the answer is pH=7

100 degrees, KW=10-12, then the NaOH concentration is 10-3mol L, and the H2SO4 concentration is 10-4mol L; If pH=7 is required, then the OH- concentration is required to be 10-5mol L, then the volume of NaOH is Xl, the volume of H2SO4 is YL, X*10-3-Y*10-4=(X+Y)*10-5, and the solution is obtained, 9X=Y, then the volume ratio of the two is X:Y=1:9

NaOH reacts with H2SO4 in a 2:1 relationship. Set to v1 and v2

2*(v2)*10 -4=(v1)*10 -5, then v1:v2=20:1

First of all, ph+poh=14, i.e. c(h+)c(oh-)=1 x 10 (-14) If you know this, let's talk about the following things.

For sodium hydroxide solution, the hydroxide ion concentration is , the concentration is 1 x 10 (-5)) For the barium hydroxide solution, the hydroxide ion concentration is, as explained above.

After mixing two solutions of the same volume, the hydroxide ion concentration is.

c(oh-)= mol/l

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

1 volume of pH = 9 NaOH solution and 1 volume of pH = 13 Ba(OH)2 solution after mixing. h+=

You don't seem to have given the prerequisites, the value of kW is different at different temperatures. kW is 14 at room temperature, but kW is 12 at 100°C

Upstairs miscalculated, calculated oh-, divide it by kw.

There is H+ 20*10-3*10-5=2*10-7mol in the original solution, the amount of OH- substance added is , OH-excess, H+ is consumed, the amount of remaining OH- substance is 2*10-7mol, and the concentration of OH- is 2*10-7 20*10-3=10-5 (the volume is too small, and the volume change is negligible); poh = 5, ph = 14 - poh = 14 - 5 = 9, so the pH value is 9.