In a Neutral Solution the Concentration of _____

Acids and Bases

There are several definitions of what constitutes an acid:

The Arrhenius definition: By the 1884 definition of Svante Arrhenius (Sweden), an acid is a textile that can release a proton or hydrogen ion (H⁺).

The Lewis definition: A Lewis acid is one that can accept a pair of electrons and form a coordinate covalent bond.

The Brønsted-Lowry definition: Brønsted argued that all acid-base reactions involve the transfer of an H⁺
ion, or proton. Water reacts with itself, for example, by transferring an H⁺
ion from 1 molecule to some other to form an H₃O⁺
ion and an OH^–
ion. According to this theory, an acid is a “proton donor” and a base is a “proton acceptor.”

For now we will stick with the Brønsted-Lowry definition. And nosotros will now tie the concept of acids and bases into equilibrium:

Permit’s start by looking at the Ionization of Water:

2H₂O(l) <=> H_threeO⁺(aq) + OH^–(aq)

This reaction does fit the Brønsted-Lowry definition since one water molecule is acting as an acid “altruistic” a proton and the other is acting as a base “accepting” the proton:

If we now write this reaction into an equilibrium expression:

Just remember ane of our rules regarding the writing of equilibrium expressions states that we do not include solvents (liquids) as their concentration stays pretty abiding during a reaction and so the h2o portion of the equation is essentially equal to ane and thus “disappears”.

The resulting equation is given a special designation K_W
called the ionization constant of water:

The value of this constant at 25^oC is 1.0 ten 10^-14. Every bit y’all tin come across from the very small value of M_West, water is not very dissociated at room temperature and our assumption that the concentration of water being abiding is a good ane.

For pure h2o, the values of [H₃O⁺
] and [OH^–
] are equal and thus their values are both 1.0 10 10^-seven. If there is an equal amount of acid and base of operations nowadays in whatever solution, the solution is called “neutral” and the pH of the solution is 7.

If the acrid concentration is greater than the base concentration, the solution is acidic and the pH volition exist less than seven. If the base concentration is greater than the acrid concentration, the solution is basic and the pH will be greater than 7.

The pH Scale:

The
pH
of a solution is defined every bit the
negative log₁₀
[H⁺]

(NOTE: H⁺
and H_threeO⁺
are interchangeable ways to describe the presence of protons in solution)

The
pOH
of a solution is defined equally the
negative log_ten
[OH¯]

Conversely,

 [H₃O⁺] = 10^-pH
and [OH¯] = 10^-pOH

The pH of a neutral solution is therefore 7.  (-log[1 x 10^-7] = 7)

The sum of the pH and pOH must always equal 14.  This is because the –log G_W
= -log(1 x x^-14) = 14.

Concentration (mol/L)
[H 3 O + ]	pH	[OH – ]	pOH
1		1 x 10-xiv	14
1 x ten-i	i	ane x 10-xiii	13
1 x 10-two	two	1 x ten-12	12
1 x 10-3	iii	1 x ten-eleven	xi		Acidic Solution
1 10 x-four	4	1 x 10-ten	x
i 10 ten-five	v	i x 10-9	9
1 x 10-vi	six	1 x 10-8	8
i x 10-7	vii	1 x ten-vii	7		Neutral Solution
1 x 10-viii	8	1 x 10-6	6
1 10 ten-9	ix	1 x 10-v	5
i 10 x-10	10	i 10 10-iv	4
i x 10-11	xi	1 x 10-three	3		Bones Solution
one ten x-12	12	i x 10-two	2
1 x 10-thirteen	thirteen	one x 10-i	1
1 x 10-14	14	1

More generally, acid dissociation constants (K_a) are of the form:

Where [HA] is the reactant acid concentration, [H+] (also sometimes represented as [H₃O⁺]) is the product acrid concentration, and [A-] is the product base concentration.

The reactants and products in an acrid-base of operations equilibrium are called “conjugates” of i another. This is because the acid and base of operations only vary by the addition or subtraction of a proton.

Instance:

H₃PO_four
+ H₂O <=> H₃O⁺
+ H₂PO₄
^–

In this example, the H_iiiPO₄
and H₂PO_four
^–
are acid and conjugate base of operations respectively, and the H₂O and H₃O⁺
are base of operations and conjugate acid respectively.

What are the conjugate bases of these acids?

What are the conjugate acids of these bases?

Answers

Potent Acids and Bases:

Now that we have defined what an acid is and how to define a solution as acidic or basic, allow’s talk about what makes an acrid strong versus weak:

This again is an equilibrium concept. Strong acids are those that profoundly dissociate in solution. In that location is a limited number of those acids, so y’all should be able to remember them (hint, hint).

The mutual acids that are most 1 hundred per centum ionized are:

HNO
_three

– nitric acid

HCl – hydrochloric acid
H
₂
And so
₄

– sulfuric acid

HClO
_iv

– perchloric acrid

HBr – hydrobromic acid
HI – hydroiodic acid

There are others that are not so common, but I wouldn’t expect you to know those unless I gave you their K_a
value. Mostly, any acid with a pKa (-logK_a) less than about -2 is said to exist a strong acid.

Values of M

_a


for Common Acids

A potent base of operations is able to deprotonate very weak acids in an acrid-base reaction. Compounds with a pKa of more than than about xiii are chosen stiff bases. Common examples of strong bases are the hydroxides of alkali metals and alkali metal globe metals like NaOH and Ca(OH)₂.

A weak acid will take a stiff conjugate base and a strong acid will have a weak cohabit base of operations.

Example pH Calculations:

(1) What is the pH of (a) homo blood, in which the hydronium ion concentration is 4.0 x 10^-viii
K; (b) 0.020 Thousand HCl(aq); (c) 0.040 One thousand KOH (aq); (d) household ammonia in which the [OH¯]= 3 x 10^-3
Thou; (east) six.0 10 ten^-5M HClO₄(aq)

Answers:
(a) pH = -log(four.0 ten 10^-8
M ) = 7.4; (b) pH = -log(0.020 Chiliad ) = 1.vii; (c) pOH = -log(0.040) = ane.four and then pH = 14-i.iv = 12.6; (d) pOH = -log(3 10 10^-3
M) = 2.5 so pH = 14-two.5 = 11.5; (east) pH = -log(half-dozen.0 x 10^-5? M ) = four.2

(2) Find the hydronium ion concentration in a solution with pH = 4.83

Reply:
[H+] = 10^-pH
= 10^-iv.83
=1.5 10 10^-five
M

(3)The pH of a solution of Ba(OH)_ii
10.66 at 25ºC.  What is the hydroxide ion concentration in the solution?  If the solution volume is 125 mL, how many grams of  Ba(OH)_two
must have been dissolved?

Answer:

14 – 10.66 = 3.34= pOH        [OH^–] = ten^-pOH
= ten^-3.34
= 4.six ten ten^-four
Thousand

1 mole of Ba(OH)_ii
for every two moles of OH^–
so 4.6 x 10^-4
mol OH 10 1 Ba(OH)₂/2 OH^–
= 2.3 x 10^-4Grand Ba(OH)₂

0.125 50 x 2.3 x ten^-four
M Ba(OH)₂
= two.9 x 10^-5
moles Ba(OH)₂
x 171g/mol = 4.ix x 10^-3
g