About the calculation method of chemical pH comprehensive additional example questions Thank you

The amount of OH-1 substance is:

The amount of molh+ substance is:

The remaining after the mol reaction is H+, and the amount of its substance is.

mol Therefore, the concentration of H+ in the solution is.

Its pH is 22, (m+16+1)g·mol-1=m, n=, and therefore the relative atomic mass of m is.

m=23g·mol-1

is an element of sodium.

C(h+) = negative Sakura at a certain temperature of pure water to the 7th power mol l. At this temperature, kw=4x10 -14, pH is 7-LG2 and the minus 13th power of C(H+)= in a solution ionized by water mol L, C(OH-)=10 -1mol L, then the pH value of the solution may be 13-2lg2

The water is off the pie Kaizi product kw to the minus 14 power.

There are two possibilities: the solution is alkaline, pH = logc (h+) = the solution is acidic, then the oh- ionized by water is also the coarse minus thirteenth power mol l with kw divided by c (oh-) so ph = 1

This question can be hypothetically used and does not have to be calculated for every option.

A is wrong, if kw

1*10 -14, then the pH of the neutral solution will be 7b wrong, pod = 2

pod=-lgc(oh-）

The kw of heavy water is not.

Then pd is not equal to 13

c pair, c(d+) = pd

d False. n(naod)=n(dcl)=25mmol, just completely neutralized, the solution is neutral, and pd is definitely not equal to 12d1:10 x so it is right).

It can be assumed that the law ,。。 Armor... Second. ph

c(h) ratio.

。Armor... Second. ph

c(h) ratio.

1：100=（1：10^x)

The title of the supplementary question is not clear.

HCl has hydrogen ions105

NaOH has hydrogen ions of 10-12

Then the ionic product of the water is 10 -14

So NaOH has hydroxide 10-2

Neutralize it. Hydroxide is left 10 -2 minus 10 -5

The difference is 3 orders of magnitude.

It can be approximated as 0

So hydroxide is 10 -2

So the ph is still 12

I kind of forgot about this part.

Probably not necessarily.

Let's see what others have to say.

Let HCl be 1L, then there will be 10 (-5)mol of hydrogen ions

The hydroxide group consists of 10 (-2) mol, and after mixing, it is 10 (-2)-10 (-5) 2, which is approximately equal to 10 (-2) 2, then pH=12-lg2

Let the volumes be V, N(H+)=V*10 (-5) in HCl, and N(Oh-)=V*10 (-2) in NaOH

Obviously, N(OH-) > N(H+), so it is alkaline when mixed.

The concentration of OH- is c(oh-)=[v*10 (-2)-v*10 (-5)] (2v)=5*10 (-3).

c(h+)=2*10^(-12)

So ph=-lgc(h+).

12-lg2=

Note: (1) 10 (a) means 10 to the power a!

2) The concentration of oh- is c(oh-)=[v*10 (-2)-v*10 (-5)] (2v)=5*10 (-3) The approximate idea is used in the calculation process, because.

10 (-2) and v*10 (-5) differ by a factor of 1000.

In the hydrochloric acid of pH = 5, the concentration of H+ is 10 to the minus 5 power, the hydroxide concentration in the sodium hydroxide solution of pH = 12 is the minus 2 power of 10, and the difference is three orders of magnitude after the equal volume is mixed, so the hydroxide of hydrochloric acid and hydroxide reaction is negligible, and the hydroxide concentration in the mixed solution is 5 times 10 to the negative power, and the negative logarithm of ten is taken, pH =.

The pH of 3mol L of H ions in solution is about pH 0, and the concentration of H is 1mol L

pH 1, h concentration.

pH 2, h concentration.

pH 3, h concentration.

pH 4, h concentration.

And so on. pH value 14, OH concentration 1mol LPH value 13, OH concentration.

pH 12, OH concentration.

pH 11, OH concentration.

pH 10, OH concentration.

And so on.

The pH value has its limitations, it can only be 0 to 14, for example, a solution with a hydrogen ion concentration greater than 1mol L cannot be expressed by pH.

The pH value is converted into the concentration of hydrogen ions or hydroxide ions, the amount consumed by the neutralization reaction after the solution is mixed, and the dilution caused by the increase in the mixing volume is calculated, and the concentration of hydrogen ions or hydroxide ions in the final solution is calculated, and the pH value of the mixed solution can be further calculated.

For the final solution to be pH = 7, the equivalence of hydrogen ions and hydroxide ions must be equal to calculate the corresponding solution volume ratio.

1.Choose D because as long as it is a strong acid, then the pH will increase by 1 if it is diluted 10 times, so whether it is a strong acid or a strong acid, as long as the pH = 3 at the beginning, then the pH will increase by 1 after 10 times dilution (but it must be less than 7). However, when the weak acid is diluted, if the equilibrium does not move, then its pH also increases by 1, but when the weak acid solution is diluted, it promotes the ionization of the weak electrolyte, so the weak acid will also ionize, so that the concentration of hydrogen ions is greater, so the increase in pH should be less than 1 does not meet the topic.

The concentration is 10 -13

First of all, it is an excess of acid, so the hydrogen ion concentration of the solution can be calculated by the following formula: (The second question, what is the hydrogen ion concentration of water ionization, then the hydrogen ions in the solution cannot be calculated at this time.) Because hydrogen ions are not only ionized by water, hydrochloric acid is also ionized.

One of the implicit conditions in the question is that there are as many hydrogen ions as hydroxide ions in water ionization. Since you can't start with hydrogen ions, you can start with hydroxide ions. Then think again, the hydroxide ions in the solution are completely ionized by water, because as long as it is in an aqueous solution, there is always a product of the concentration of hydrogen ions and hydroxide ions is 10 -14 (the ion product of water), and the concentration of hydrogen ions in the solution is 10 -13.

These hydroxide ions are completely ionized by water, and the synthesis of "the concentration of hydride and hydrogen ions in water ionization is the same", and the hydrogen ionization of water ionization is also 10 to the minus 13th power.