How do you find the pH of a solution of two strong acids?

Answer 1

A strong base is defined as one that dissociates completely in water (see the Related Questions link to the left for a list of the strong bases). That means for every mole of base added, one mole of free OH- is present in the solution. The pH of solution is defined this way:

pH = -log10[H3O+]

or in English, the pH is equal to the negative logarithm (in base 10) of the concentration of H3O+ in the solution (the concentration must be in units of Molarity (M), which is moles per liter).

To use this, we need to know the concentration of H3O+. But to do this, first we must find the concentration of OH- and we can then use that to find the concentration of H3O+.

Because a strong base dissociates completely in water, the number of moles of base added gives the number of moles of OH- present in the solution. There is one complication that you don't have to worry about with strong acids. Two strong bases have 2 OH-'s in their formula, like Ba(OH)2 and Sr(OH)2. These bases release TWICE as many moles of OH- as the number of moles of base added. I will show this in an example below, but first we need to know how to get the concentration of H3O+ from the concentration of OH-! They are related in this way:

[H3O+] * [OH-] = 1.0 * 10-14

Or in English, the concentration of H3O+ times the concentration of OH- is ALWAYS equal to 1*10-14 in water. So once we know the concentration of OH-, we can easily find the concentration of H3O+ this way:

[H3O+] = (1.0 * 10-14) ÷ [OH-]

Then we use the definition of the pH above. Here are a few examples:

* If you have a solution of 0.1 M lithium hydroxide (LiOH), what is the pH? The concentration of OH- is the same as the concentration of LiOH, 0.1 M.

So the concentration of H3O+ is:

[H3O+] = (1 * 10-14) ÷ 0.1 = 1 * 10-13 M

The pH is then:

pH = -log (1 * 10-13) = 13

* If you have a solution of 1.0 M potassium hydroxide (KOH), what is the pH? Just as above, he concentration of OH- is the same as the concentration of KOH, 1.0 M.

So the concentration of H3O+ is:

[H3O+] = (1.0 * 10-14) ÷ 1.0 = 1.0 * 10-14 M

The pH is then:

pH = -log (1.0 * 10-14) = 14

* Let's try a twist. If you have a solution of 0.001 M Ba(OH)2, what is the pH?

Now the concentration of OH- is twice the concentration of Ba(OH)2, so we have:

[OH-] = 2 * 0.001 M = .002 M

Now we proceed just as before. So the concentration of H3O+ is:

[H3O+] = (1.0 * 10-14) ÷ 0.002 = 5.0 * 10-12 M

The pH is then:

pH = -log (5.0 * 10-12) = 11.3 Strong acids and bases have all of the dissolved material completely ionized. The concentration of a monoprotic acid is equal to the concentration of hydrogen ions. The concentration of a monobasic alkali is equal to the concentration of hydroxide ions. The actual concentration of hydrogen ions from pure water is on the order of concentration of E-7 molar, so any concentration of a strong acid or base over E-5 molar completely swamps the comparatively tiny amount of ion from the ionization of water.

One bobble point that comes up with newbies is: What is H3O+ and why do we use it.

Hydrogen ions don't float around free in solution, they associate with water molecules at the Oxygen side. The slight negative charge of the polar bonds attracts the positive Hydrogen atom in a weak temporary bond that allows the ions to move from molecule to molecule. This effect is often found when Hydrogen is part of a molecule and results in things like holding DNA strands together, color separation in color chromatography, Gas Chromatograph column separations,... Hence the name: Hydrogen Bonding

Answer 2

A strong acid is defined as one that dissociates completely in water (see the Related Questions link to the left for a list of the strong acids). That means for every mole of acid added, one mole of free H+ (or equivalently, H3O+) is present in the solution. The pH of solution is defined this way:

pH = -log10 [H3O+]

or in English, the pH is equal to the negative logarithm (in base 10) of the concentration of H3O+ in the solution (the concentration must be in units of Molarity (M), which is moles per liter).

So to find the pH of a strong acid solution all you need to know is the concentration of the solution. Let me give a couple of examples.

• If you have a 0.01 M solution of hydrochloric acid, HCl, what is the pH?

pH = -log (0.01) = 2

• If you have a 0.007 M solution of sulfuric acid, H2SO4, what is the pH?

Well, this is special because sulfuric acid is a (quite) strong diprotic acid in diluted solutions (below 0.1 M). Therefore the concentration of H3O+ is twice the molarity of H2SO4 because this holds 2 protons (H+) per mole.

So in a 0.007 M solution of sulfuric acid, H2SO4, the concentration of H3O+ is:

0.007 × 2 = 0.014 M and

pH = -log (0.014) = 1.85

(i.s.o. pH = 2.15, when only one of the two H's in H2SO4 were released completely. However, the second H is also to be considered as strong in dilute solution, its pKa is 1.92, implicating that more than half of these second H's is protolysed at pH below 1.92, according to the Henderson-Hasselbalch equation)

• If you have a 1.0 M solution of nitric acid, HNO3, what is the pH?

pH = -log (1.0) = 0

Yes, the pH can be zero. In fact it can be negative!

• If you have a 5.0 M solution of HCl, what is the pH?

pH = -log (5.0) = -0.7

NOTE: This is NOT the same as with a WEAK acid solution. To find out how to find the pH of a weak acid, see the Related Questions to the left of this answer (and also to find out if you have a strong or weak acid).

Answer 3

NOTE: This is NOT the same as finding the pH of a STRONG acid solution. See the Related Questions links at the end of this answer to find out the difference between a strong acid and a weak acid, and if you have a strong acid, follow the link to find the pH of a strong acid solution (it is much easier!)

This is long and boring so i will simplify it. Usually you use a universal indicator and a ph scale. You can get those things from shops like bunnings (soil ph level)

To determine the pH of a weak acid solution you must know two things: you need the concentration of the acid in the solution, and you need the Ka of the acid (or equivalently the pKa, which you can use to calculate the Ka).

First, we must write the equation of the acid dissociation in water. Let's use a generic acid that we'll call 'HA.' When it dissociates, it will form H+ and A- (A- is called the conjugate base of the acid HA).

HA + H2O --> H3O+ + A-

Note: When describing acids, some people use H+ and some people use H3O+, but they are basically equivalent. It is a bit more accurate to use H3O+ because that is what is actually present in the solution, so I will use it here.

Because this is a weak acid, this reaction will go to some equilibrium value, and this will be described by the equilibrium constant, Ka. The equilibrium product for this reaction is found by taking the concentration of the products and dividing them by the concentration of the reactants (with the concentration of each specie raised to its respective coefficient in the balanced reaction). So at equilibrium, we have:

Ka = [H3O+] [A-] ÷ [HA] - where the square brackets mean concentration (for instance 'N]' means 'the concentration of N').

Note: H2O is NOT included in the equilibrium product even though it is a reactant because its concentration in dilute solutions is not changing more than 0.1%]

If you are given the pKa instead of the Ka, use this formula to find the Ka:

pKa = -log10 Ka or inversely Ka = 10-pKa

OK -- now how do we solve? To do this, we must set up what is sometimes called an 'ICE' chart, for Initial-Change-End. Because of the formating constraints, it will be hard to show that here correctly, but I will describe it. Write the balance equation, and underneath it we're going to write three rows of information

In the first row, called 'Initial,' write the concentration of HA, H3O+ and A-initially (in other words before equilibration). Well, that's easy. Initially, we have zero concentration of H3O+(*) (cf. note at the end) and zero concentration of A-, and the amount of HA is however much was specified in the question (in units of molar (M), or moles per litre). To keep this explanation general, let's call the number which is the starting concentration of HA "[HA]initial." This is our "initial" row of data.

In the 2nd row, the "Change" row, we're going to mark the change in the concentration of each specie. Well, we don't know that! So we're going to use a variable, and let's call it "x." Now, the only way to make H3O+ and A- is from an HA molecule that dissociated. So we know that however much A- is formed, we must have made exactly as much H3O+ at the same time. And we also know from the stoichiometry of the balanced reaction, that for each mole of A- formed, one mole of HA was reacted.

So, let's say that the change in [H3O+] increase by 'x' (note that it has to increase -- it started at 0 and you can't have a negative concentration!), then the [A-] must increase by exactly the same amount, so also 'x'. Now that means that [HA] decreased by the same amount... 'x'... except because it decreased, we make it negative, and write '-x'.

Now we'll fill in the 3rd row, the "End" row. To find this, we just added up the first two rows. So, in the end,

[H3O+] = 0 + x = x (*) (cf. note 1 at the end)

[A-] = 0 + x = x

The end value of [HA] is going to be whatever it started at minus x. So that can be written as: [HA]end = [HA]initial - x

Now we can solve.

Ka = [H3O+] [A-] ÷ [HA] = x * x ÷ ( [HA]initial - x )

To find the pH, we want to solve for [H3O+], which happens

to equal 'x'. (*) (cf. note 1 at the end)

Rearranging, we have:

x2 - ( Ka = ([HA]initial - x) ) = 0

If the acid is fairly weak and/or not too diluted(**) (cf. note 2 at the end), we can say that not much will dissociate, and can make the following approximation to simplify the math:

[HA]initial - x = [HA]initial

So now, using that, we can write:

x2 - (Ka*[HA]initial) = 0 or x2 = (Ka*[HA]initial)

And we know what both Ka and [HA]initial are from the beginning, so just multiply them together and take the square root to solve for 'x'.

The last step is to find the pH. The pH is defined like this: pH = -log10[H3O+]

It turns out that the value of 'x' is the same as [H3O+] because of the way we set up the problem. So to find the pH, just take the negative of the logarithm (base 10) of 'x'.

pH = -log10 (x)

And you're done!

Note 1:

(*) Actually this is not the full truth: Water itself initially contains 10-7 protons.

Thus exactly [H3O+] = 10-7 + x which is approximately equal to x if x>>10-7, for example x= [H3O+]calculated=10-6 or bigger.

So the outcome of this calculation is still usable if -log10 (x) = [pH]calculated =< 6.

If not, the pH can be fairly approximated to be between 6 and 7,because it should clearly never exceed value 7 (Remember: Water itself initially contains 10-7 protons!)

Note 2:

(**) Otherwise one has to solve the quadratic equation:

x2 + Ka*x - Ka*[HA]initial = 0

resulting in:

x = 0.5 * [ - Ka + SqRoot( Ka2 + 4*Ka*[HA]initial) ]

Answer 4

See these two questions:

[http://wiki.answers.com/Q/How_do_you_find_the_pH_of_a_strong_acid_solution

How do you find the pH of a strong acid solution?]

and

[http://wiki.answers.com/Q/How_do_you_find_the_pH_of_a_strong_base_solution

How do you find the pH of a strong base solution?]

Answer 5

I was going to say with pH check but then I ran into this: "Do not confuse the term acid strength with pH. The strength of an acid has to do with the percentage of the initial number of acid molecules that are ionized. If a higher percentage of the original acid molecules are ionized, and therefore, donated as hydrated protons (hydronium ions) then the acid will be stonger. Strong acids are Hydrochloric (HCl(aq)), Hydrobromic (HBr(aq)), Nitric (HNO3), Sulfuric (H2SO4), and Perchloric (HClO4) acids. In each of these molecular acids the percentage of ionization is almost 100%" Copy and paste this link for more information than you probably want.

http://members.aol.com/logan20/acid_str.html
The strength of an acid is determined by how completely it ionizes when dissolved in water.

Answer 6

You need to put a Universal Indicator (dark green liquid substance) into the 2 strong acids. There is a pH color code and is determined by what color it is. Under 7pH is acidic (yellow-red), and over 7pH is basic (dark green-purple), and 7 is neutral (green).

Added:

In dilute solution it is allowed to add up both concentrations of the two strong acids and then calculate as it were one (total) strong acid.

(Not allowed for one strong and one or more weak acids or sulfuric acid ! )

Answer 7

pH = -log [H+]

Confused? This rule applies to bases as well, but in a more complicated way because the concentration of [H+] isn't apparent. As it turns out, there are still H+ ions present because the aqueous part of solution is constantly exchanging H+ and OH- in equilibrium (autoionization).

In order to use the pH formula, we must convert the concentration of base to the concentration of acid using the equilibrium expression for autoionization of water. This equilibrium constant, Kw, equals 10-14. The pH formula includes a negative only because we like positive numbers for convenience (log (10-14) is negative). Anyway, to solve for [H+] of a basic solution, you simply need to divide 10-14 by the number of moles of hydroxide ions.

Example:
Consider a basic solution of [OH-] = 0.050 M
Kw = [H+]*[OH-] = 10-14
[H+][0.050 M] = 10-14
[H+] = 10-14/[0.050 M] = 2.0 * 10-13 M

Now that we know the concentration of H+ in the basic solution, we need only to take -log of that number of moles.

pH = -log [H+] = -log [2.0 * 10-13 M]
pH = 12.7

Answer 8

To find out the PH of an unknown solution you can either place blue/or red litmus paper into the solution and observe for a few minutes to see the colour change on the paper. If the paper turns purple then the solution is known as an Alkaline but if the paper turns orange/red then that means the solution is an Acid. Another way to discover the PH of a solution is to use Universal Indicator along with a PH scale chart. It is simluar to the Litmus paper observation. If the solution is a shade of red then check with the PH chart to find out the exact PH number of the solution, ideally, the more deeper the red, the lower PH that the solution will have, this will apply the same if the solution turns into a shade of purple. Ther darker the purple, the higher the PH will be.

Answer 9

1.0 (one)