Chemical pH calculation, how to calculate pH from chemical formula?

Monobasic acid, indicating that the acid is a strong acid.

Monobasic base indicates that the acid is a weak base.

The mixture is because of strong acids, weak alkalis, salts and weak bases.

Strong acids and weak alkali salts are acidic, weak alkalis are alkaline, and the concentrations of the two are different, and it is possible to neutralize, pH = 7

Because the pH value of the two is different, it can be simply understood that the pH is 1 to 14, while the acid pH is only 2, and 1 is one level worse. The alkali pH is 12, and 14 is two levels different, one level is 10 times the difference, which is equivalent to the same volume and concentration of pH2 acid and pH13 alkali is neutralized, reaching 7, and the alkali is obviously one level worse, so only rely on concentration to make up.

Monobasic acids are mixed with monobases.

Therefore, the monobasic acid is a strong acid, and the monobasic base is a weak base.

The pH after mixing will not be pH7

Equal volume, then you might as well set the volume to 1L, and the volume after mixing is roughly regarded as 2L acid, which is recorded as HA alkali and recorded as BOH

The excess alkali after mixing is calculated, and the mixture solution is c(ba)=c(boh)=according to the degree of weak alkali dissociation is 10%, and kb is about, and poh=pkb-lgcb ca= is calculated

Calculate that pH=, and apparently pH cannot be 7

The former is a strong acid, and the latter is a weak base. Weak alkali ionization 10%.

But where is the answer? It's also a master to be able to mix 7.

It's impossible that you're reading the wrong question.

Calculation of pH:

H+ concentration is often used to indicate the acidity and alkalinity of a solution.

When [H+] is less than 1mol·l-1, the negative logarithm of hydrogen ion concentration is commonly used for ease of use.

i.e. -lg[h+] to indicate the acidity of the solution and is called ph, i.e. ph = -lg[h+].

Detailed explanation: 1. [H+]·oh-]=kw in aqueous solution of any substance, kw=1 10-14In pure water, [H+]=OH-]=10-7mol·L-1, then pH= -Lg[H+]=7

In other neutral solutions, [H+]=OH-]=10-7mol·L-1, pH is also 7; [H+] in acidic solution

2. [OH-], its pH 7;Alkaline solution [H+] OH-], its pH 7The negative logarithm of the concentration of hydroxide ions can also be expressed as POH, then the pH + POH = 14 and pH = 14 - POH of the solutionThe key to calculating the pH of a solution is to correctly find the [H+] of each solution.

The specific calculation is as follows:

Example 1 Calculate the pH of hydrochloric acid solution;

Hydrochloric acid is a strong electrolyte.

Fully ionized in water [h+] = ;

Answer: The pH of this solution is 2;

Example 2 Calculate c = acetic acid.

Solution (degree of ionization.

ph; Acetic acid is a weak electrolyte.

partially ionized in water;

h+]=c=；

ph= -lg[h+]=lg ；

Answer: The pH of the solution is;

Example 3 Calculate C(NaOH) = sodium hydroxide.

the pH of the solution;

The solution NaOH is a strong electrolyte that is fully ionized in water;

oh-]=；

ph= -lg[h+]=lg10-13=13；

ph＝-lg［h＋］

pH -lg H, H is the concentration of H e.g.: HCl where H brings into the formula: Ph -Lg H Lg so his pH 1 again as:

CH3COOH, where H is simply or brought into the formula: PH -LG H LG Hengwu so his pH 2 is like this, depending on H, which is the concentration of H.

How to calculate the pH value:

How to calculate the pH of a single solution:

1. Strong acid. Cmol·l 1hna strong acid solution, c(h) ncmol·l 1 ph lgnc.

2. Strong alkali. Cmol·L 1b(OH)N strong alkali solution, C(OH) ncmol·l 1, N(H) mol·l 1 pH 14 lg nc. <>

1. Question: pH=1 then [H+]= (mol L) After mixing, [H+]= (mol L).

pH=4, then [H+]=1*10 -4 (mol L) after mixing[H+]=1*10 -4) 2=5*10 -5 (mol L).

In this case, the concentration of hydrogen ions is: [H+]= (mol l).

then after mixing, pH=

2. The concentration of hydrogen ions in pH=10 NaOH solution is [H+]=1*10 -10 (mol L).

After equal volume mixing, [H+]=1*10 -10 2=5*10 -11 (mol l).

The hydrogen ion concentration of the pH=14 NaOH solution is [H+]=1*10 -14 (mol L).

After equal volume mixing, [H+]=1*10 -14) 2=5*10 -15 (mol l).

The sum of the hydrogen ion concentrations of the mixed solution [H+] = 5*10 -11) + 5*10 -15) = (mol l).

then after mixing, pH=

3. Hydrochloric acid with pH = 1, which is [H+] = (mol L).

Sodium hydroxide at pH = 12, POH = 14-pH = 2. i.e. [oh-] = (mol l).

hcl + naoh = nacl + h2o

After the disadvantage of equal volume mixing, there is a surplus of hydrochloric acid, and its hydrogen ion concentration is: [H+]= (mol L).

So the pH of the mixed solution = -lg (

The pH of a strong acid solution is directly obtained from C(H+).The strong alkali solution is first found by C(OH-) and then by pH.

How to use an acidity meter (pH meter).

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.

The amount of NAOH is, the amount of HAC is, the composition of the mixed solution is equivalent to: HAC and NAAC, this buffer solution, the pH calculation formula is: PH=PKA+LG(CBCA)=

This question is the pH value of the buffer calculation, after the reaction can be considered that the solution contains:,, by the formula: pH = pka-lg(HAC Naac) =

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 =.

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.