Acids and bases are an essential components of numerous chemical reactions and systems. To understand chemistry, one must know acid-base chemistry — and the two constantly go together.

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Any chemical substance deserve to either be acidic (strongly or weakly), neutral, or basic (strongly or weakly). Below are a few examples that acids, bases and neutral compounds along a continuum, from solid acid to strong base.


We will encounter other, an ext subtle meanings of acids and bases later, but for now, a an excellent working definition that is beneficial most of the moment is:

Acids room H+ (proton) donors

Bases room OH- (hydroxyl ion) donors

Acids are substances the either directly or indirect raise the concentration of H+ ions (protons) in solution. Bases either straight or indirectly raise the concentration of OH- (hydroxyl ions) in solution.

You will desire to memorize the strongest acids and also bases (table below).

solid acids

HCl, HBr, HI– the halogenic acids

H2SO4 – sulfuric acid

HNO3 – nitric acid

HClO4 – perchloric acid

strong bases

KOH – potassium hydroxide

NaOH – sodium hydroxide

Ba(OH)2 – barium hydroxide

Ca(OH)2 – calcium hydroxide



In the sense of acids and also bases, "strong" method that the compound totally dissociates in water. Weak acids and also bases just dissociate partly, or are insoluble.

HCl, hydrochloric acid*, is a strong acid—one of the strongest. That dissociates totally into a chloride ion and a totally free proton (from right here on, I"ll speak to a +1 hydrogen ion a proton, since that"s what the is).

HCl → H+ + Cl-

It"s the cost-free proton that provides a strong acid different from a neutral compound. If one mole of HCl is liquified into one liter that water, the concentration the H+ will be 1M because HCl dissociates totally (it is a strong electrolyte - that"s the reason for the solitary arrow in the dissociation equation). The very same is true the the other acids top top the strong acid list. Us say that HCl "donates" the proton come the solution—very willingly in this case.

*HCl is a gas at room temperature and also pressure.


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Sodium hydroxide, NaOH, is among the most typically used solid bases. It, too, dissolves totally in water:

NaOH → Na+ + OH-

It develops a salt cation and also a hydroxyl anion, OH-. NaOH is stated to "donate" a hydroxyl ion come the solution.

Sodium hydroxide is a common component of drainpipe cleaners, and is likewise known through the common name, "lye". The is a white solid in ~ room temperature and pressure. Base compounds are likewise called alkali compounds native time to time.

Neutral Compounds

Water is the prototypical neutral compound. Pure water is what we specify neutral come be. The is neither acidic nor basic. A closer look later will display us the it"s in reality both a weak acid and a weak base. We recognize that, uneven the H-atom in HCl, the H-atoms in H2O space bound covalently,

and don"t come off easily. Water is no an ionic compound, however it still dissociates. At any given time in a maker of water, part molecules might be found undergoing the dissociation, known as the auto-ionization that water:

H2O ⇌ H+ + OH-

Neutralization reactions

A vital feature that acids and also bases is the they deserve to neutralize one-another. Think about the reaction between the strong acid HCl and also the strong base NaOH:

HCl + NaOH → NaCl + H2O

This is usual of a neutralization reaction. The products of neutralization room a salt consisting of the liberated cation the the base and the anion of the acid, and also water. Because simple salts don"t save ionizable protons or hydroxyl groups, they are neutral in solution, and of course water is neutral. A solution of NaCl in water, because that example, is neutral.

A Closer Look at the Auto ionization the Water

We provided above that at any kind of given time, some water molecules space dissociated right into OH- and H+ follow to the equation

H2O ⇌ H+ + OH-

Often, the equation is written as

H2O + H2O ⇌ H3O+ + OH-

where H3O+ is referred to as the hydronium ion. We regularly write acid base equations utilizing the hydronium ion due to the fact that it an ext accurately mirrors what in reality happens within an aqueous solution, but the easier equation is commonly perfectly sufficient for our purposes. A complimentary proton is never exactly "free" in water. It constantly associates, if just fleetingly, v one or much more water molecules, yet is quiet able to move quickly through the liquid.

We have the right to write the equilibrium consistent expression for this reaction like this:


Now in a neutral systems = , for this reason

= = 2 = 1.0 x 10-14,

thus in a neutral solution, = 1.0 x 10-7.

Now let"s execute something that may seem odd, ... And also I"ll explain later: Let"s take the an adverse of the base-10 log in of both sides of that last equation:

-log10 2 = -2 log10 = -log10 (1.0 x 10-14),

which gives

= 7.

We contact this number, the an adverse of the base-ten log in of the proton concentration, the pH the the solution. The pH that pure water is 7, and also we describe pH = 7 together the neutral pH.

The meaning of "p" in chemistry

In chemistry, the letter p means "take the negative base-ten log," -log10, because that example:

pH = -log

pOH = -log

pKw = -log

Although we might use either pH or pOH to characterize the acidity or basicity of a solution, we use pH through convention. The reason we use the pH range — a logarithmic range — at all is the the variety of possible proton concentrations is very broad, from nearly 10-14 M as much as 10 M or more. ~ above the pH scale a very acidic solution could have a pH the 1 if a very simple solution might have a pH the 14. These are controllable numbers, if we can just remember whereby they came from. At any type of rate, the pH range is entrenched in how we perform chemistry today.

Also note that a pH 4 solution has actually 10 time the proton concentration as a pH 5 solution. That"s how a logarithmic range works.

pH 7 method basic.

pH = 7 means neutral.

Relationship between pH and also pOH

A tiny algebra gives us a comfortable relation: notification that due to the fact that in aqueous solution, = 1.0 x 10-14, we can always calculate if we know , and also vice-versa. We understand that if we take the an unfavorable base-10 log in of the Kw expression, we get:

-log ( ) = -log(1.0 x 10-14)

Using among the regulations of logs, log(ab) = log(a) + log(b), we get

-log - log = -log(1.0 x 10-14)

Now converting the -logs to "p" notation we acquire a really useful relationship in between pH and pOH, one the you should memorize:

pH + pOH = 14

pH + pOH = 14

Example 1

Calculate the pH and pOH the a 0.001M equipment of HBr.

Solution: The dissociation reaction is HBr → H+ + Br-. Since this is a solid acid, we assume that it dissociates completely, therefore the concentration that H+ in solution will be = 0.001 M.

$$ eginalign pH &= -log(0.001) = 3 \ pOH &= 14 - pH = 11 endalign$$

Example 2

Calculate the pH and pOH of a 0.00015M solution of KOH.

Solution: The dissociation reaction is KOH → K+ + OH-. Since this is a solid base, us assume the it dissociates completely, so that the concentration the OH- in equipment will it is in = 0.00015.

$$ eginalign pOH &= -log(0.00015) = 3.82 \ pH &= 14 - pOH = 10.2 endalign$$

Weak Acids and also Bases

Strong acids and also bases have a lot of street cred. Lock eat through things, and also that"s cool, however weak acids and also bases are the organization end that chemistry, particularly in organic systems. Weak acids and bases space a little more daunting to address because castle don"t dissociate totally in solution. That way we need to be mindful of equilibria in i m sorry acids and also bases aren"t totally dissolved.

The proton concentration in a 1.0 M systems of the weak acid acetic acid (CH3COOH), because that example, will certainly be much lower than 1.0 M (what we would intend from complete dissociation come CH3COO- and also H+) because the H+ doesn"t detach as conveniently as the does from a strong acid. The anion CH3COO- is well-known as the acetate ion.

Let"s do an instance with acetic acid: calculate the pH the a 0.1 M equipment of acetic acid.

Carboxylic acids

Acetic mountain (left) is a weak organic acid recognized as a carboxylic acid. Over there are numerous such essential acids, differing by the identification of what is the CH3 team in the figure. If the is one H atom, for example, the acid is formic acid, the acid that gives red ant bites your sting.

The carboxyl group, -COOH, hold the acidic proton, the one the ionizes reasonably easily, despite not nearly as conveniently as the proton of a solid acid.



Example 3

Dissociation of acetic acid, a weak acid

The dissociation equilibrium is


(note the twin arrow this time), with:


Here the equilibrium consistent is composed as Ka (short for "K-acid"). These and the Kb"s of weak bases have been measure up for virtually any weak mountain or basic you will certainly encounter. You have the right to download a table of lock here.

Now in a 0.1 M solution of acetic mountain (often abbreviation CH3COOH), very tiny CH3COOH will certainly actually dissociate because the equilibrium constant is for this reason small. Look again at the Ka expression. In order for that portion to be so small, the denominator should be rather a little bit bigger 보다 the numerator.

If x mole of CH3COOH dissociate, climate there will certainly be 0.1-x moles of undissociated CH3COOH left in equipment at any time. Noting that the concentrations of acetate ion and also protons should be the exact same from the stoichiometry that the dissociation reaction, we have


Multiplying both sides of this equation by the denominator top top the left and gathering terms on one side, we gain this quadratic equation:

x2 + 1.614 x 10-5 x - 1.614 x 10-5 = 0.

This deserve to be addressed by perfect the square or by utilizing a quadratic equation (Ax2 + Bx + C = 0) regimen on a calculator or computer, v A = 1, B =1.614 x 10-5, C = -1.614 x 10-5. We acquire x = 0.00252M.

Now pH = -log, for this reason pH = -log<0.00252> = 2.6. Because that comparison, the pH the a 0.1 M solution of a strong acid would be 1.

an approximation

It"s in reality not essential to fix the quadratic equation every time for weak acids. It turns out the the error in do a particular approximation is generally so little that we deserve to tolerate it, and the approximation provides life through weak acids and also bases easier. It"s outlined below.

A handy approximation

In instance 3 above, we needed to deal with the equation


which is a quadratic equation. While this isn"t also difficult, specifically if you"ve programmed your calculator to perform it, there is a practically approximation that will certainly make life easier. Look at the equation again:


Now x is the concentration the protons, which, for a weak acid, must be an extremely small.

We make the approximation the x is insignificant contrasted to 0.1 and just cross it out:


Now we simply need to main point both political parties by 0.1 and also take a square root. In this case the remedies are:

Exact pH = 2.598

with approximation pH = 2.896,

an error of around 10%.

As the dissociation constant becomes smaller sized (for weaker acids), this approximation it s okay better.

Weak bases

One the the most crucial weak bases is ammonia. That isn"t a prototypical base due to the fact that it doesn"t directly it is provided a solution with OH- ions. Right here is the aqueous equilibrium reaction:

NH3 + H2O ⇌ NH4+ + OH-

We should pause here to filter our meanings of acids and also bases, due to the fact that our previous one won"t work for ammonia. This is the Brönsted-Lowry meaning of acids and bases:

Acids are H+ (proton) donors

Bases space proton acceptors

This an interpretation covers ammonia. Take an additional look in ~ the dissociation equation. Ammonia "accepts" a proton indigenous water (note that water acts as an mountain in this case), leading indirectly to rise in hydroxyl ion concentration. The Kb for ammonia is Kb = 1.778 x 10-5. The Kb expression is


Just choose Ka values, Kbs are tabulated for numerous weak bases. Wikipedia is a pretty an excellent source because that chemical properties of every kinds, including acid-base properties.

Example 4

A weak base calculation

Calculate the pH and also pOH that a 0.001M solution of ammonia, NH3.

The dissociation reaction is NH3 + H2O ⇌ NH4+ + OH-. Because this is a weak base, we know it doesn"t dissociate completely, therefore we have to use the equilibrium continuous expression (above). If us let x be the lot of NH3 the accepts a proton native water, then (0.001 - x) is the amount of NH3 left in solution, and we have:


Here we have actually employed our approximation come get

x = = 0.001333 M

Now we can calculate the pOH:

pOH = -log(0.001333) = 2.88

... And also finally the pH the the straightforward solution:

pH = 14 - pOH = 11.12

Practice problems


Solution here


Calculate the pH that a 1.0 × 10-3 M equipment of hydrochloric acid (HCl)


HCl is a strong acid, HCl → H+ + Cl-, thus it dissociates (comes apart right into + and also - ion in water) completely.

$$ eginalign &= 1.0 imes 10^-3 ; M \ pH &= -log< H^+ > \ &= -log <10^-3> = f 3 endalign$$

pH 3 is a short pH (


Calculate the pH of a 0.9 M systems of hydrobromic mountain (HBr)


HBr is a solid acid, HBr → H+ + Br-, as such it dissociates (comes apart into + and - ion in water) completely.

$$ eginalign &= 9.0 imes 10^-1 ; M \ pH &= -log< H^+ > \ &= -log <0.9> = f 0.046 endalign$$

pH 0.046 is a very low pH (


Calculate the pH of a 2.234 × 10-6 M solution of potassium hydroxide (KOH).


KOH is a strong base, KOH → K+ + OH-, and therefore dissociates fully into negative (they hydroxide ion, OH-) and positive ion in water solution.

$$ eginalign &= 2.234 imes 10^-6 ; M \ pOH &= -log< OH^- > \ &= 5.65 \ pH &= 14 - pOH = f 8.35 endalign$$

pH 8.35 is a high pH (> 7), as such this is a basic solution.


Calculate the pH of 735 liters the a systems containing 0.34 mole of nitric acid (HNO3).


First we have to calculate the concentration the the solid acid HNO3:

$$frac0.34 ; mol ; HNO_3735 , L = 4.626 imes 10^-4 ; M$$

Now HNO3 is a solid acid, and also dissociates fully in water: HNO3 → H+ + NO3-.

$$ eginalign &= 4.626 imes 10^-4 ; M \ pH &= -log< H^+ > \ &= -log <4.626 imes 10^-4> \ &= f 3.33 endalign$$

Despite the big dilution aspect (a small variety of moles into 735 Liters), this is still quite an acidic solution, pH


Calculate the pH that a 2.0 liter systems containing 0.005 g that HCl.


First we need to calculate the concentration of the HCl. We have actually to convert 0.005g come moles, then divide by 2.0 Liters.

$$ eginalign &= frac0.005 , g ; HCl left( frac1 ; mol ; HCl37.45 , g ; HCl ight)2.0 , together ; solution \ \ &= 6.68 imes 10^-5,M endalign$$

Now usage that concentration, together with the truth that you kow HCl is a solid acid, which will certainly dissociate completely, to find the pH:

$$ eginalign &= 6.68 imes 10^-5,M \ pH &= -log< H^+ > \ &= -log <6.68 imes 10^-5> \ &= f 4.18 endalign$$


Calculate the pH and pOH of a equipment with a volume the 5.4 liters that consists of 15 g the hydrochloric mountain (HCl) and also 25 g the nitric mountain (HNO3).


There are two strong acids in this solution. We need to calculate the total number of moles the protons, then divide that by 5.4 l to acquire

$$ eginalign 15 , g ; HCl left( frac1 ; mol ; HCl36.46 , g ; HCl ight) &= 0.411 , mol \ \ 25 , g ; HNO_3 left( frac1 ; mol ; HNO_363 , g ; HNO_3 ight) &= 0.397 , mol endalign$$

So the total variety of moles is 0.411 + 0.397 mol = 0.808 mol. Now,

$$ = frac0.808 , mol ; H^+5.4 , L = 0.1495 , M$$

So the pH is

$$ eginalign pH &= -log< H^+ > \ &= -log <0.1495 , M> \ &= f 0.82 ; ; longleftarrow ; ext quite acidic \ pOH &= 14 - pH = 14 - 0.82 = 13.18 endalign$$


A swimming pool has actually a volume that 1 million liters. How countless grams the HCl would certainly you require to add to the pool to carry the pH native 7 down to 4? Assume the the volume of the HCl is negligible.


At pH 7, there will be no HCl present, presume we"re working through pure H2O. In ~ pH = 4, the variety of moles of H+ will certainly be:

$$ eginalign &= fracx ; mol ; H^+10^6 , L ; ightarrow \ \ x &= (1 imes 10^-4)(1 imes 10^6) \ &= 100 ; mol ; H^+ endalign$$

Now because that every 1 mol of H+, over there is 1 mol the HCl, for this reason the number of grams that HCl we need to reduce the pH come 4 is:

$$ eginalign 10^2 ; mol ; H^+ &left( frac1 , mol ; HCl1 , mol ; H^+ ight) left( frac36.46 , g ; HCl1 ; mol ; HCl ight) \ \ &= 3646 , g ; HCl \ &= ext 3.646 Kg HCl endalign$$


Calculate the pH and pOH that a systems that to be made by including 400 ml the water come 350 ml the a 5.0 × 10-3 M NaOH solution.


Firs we require the variety of moles that NaOH in 350 ml (0.350 L) of a 1.74 × 10-3 M solution:

$$0.350 , l left( frac5.0 imes 10^-3 ; mol ; NaOH1 ; L ight) \ \ = 1.75 imes 10^-3 ext moles NaOH$$

Now the new concentration of NaOH, after ~ dilution is:

$$frac1.75 imes 10^-3 ; mol; NaOH(0.350 + 0.400) , L = 2.33 imes 10^-2 , M$$

So the pOH and also pH are:

$$ eginalign pOH &= -log(2.33 imes 10^-2) = 1.63 \ pH &= 14 - pOH = 12.4 endalign$$


The active ingredient of aspirin is acetyl salicylic acid (C8H7O2COOH), which has actually Ka = 3.0 x 10-4. Acetyl salicylic mountain is regularly abbreviated "ASA".

calculation the pH of a systems made by dissolve 500 mg the ASA in water, and diluting it come 50 ml. Repeat the calculation, this time dissolve the ASA in 1000 ml the water. Is the pH higher or reduced in the an ext dilute solution? In which solution is the fraction of ionized ASA higher? solution

We deserve to abbreviate acetylsalicilic acid as HA, whereby A is the anion and also H is the proton the comes off. The dissociation reaction is then

HA ⇌ H+ + A-


Papaverine hydrochloride (pap-HCl) is a salt of a weak basic (papaverine) and a strong acid (HCl). That is a drug supplied as a muscle relaxant. Pap-HCl is a weak acid overall. In ~ 25˚C, a 205 mM systems of pap-Hl has a pH of 3.31. Compute the Ka of pap-HCl.


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A 0.040 M systems of a weak acid (call it HA) has actually a pH of 4.70. Calculation the Ka and pKa of the acid.