what happens to the poh of an aqueous solution of ammonia when ammonium chloride is added?

Acidic and Basic Salt Solutions

• Background Information
• Relationship between Ka and Kb of Cohabit Acid-Base Pairs
• Computing pH of a Table salt Solution

Background Information

When certain soluble salts are dissolved in water the resulting solution is not neutral.  Depending on the composition of the salt (the ions which it is made upward of) the solution will be either acidic or bones.  Soluble salts that contain anions derived from weak acids form solutions that are basic.  The anion is the conjugate base of operations of a weak acid.  For example, the acetate ion is the cohabit base of acetic acid, a weak acrid.  Therefore, a soluble acetate salt, such as sodium acetate volition release acetate ions into the solution, which a few of these will interact with water, forming unionized acerb acid and the hydroxide ion.

NaCH_iiiCOO(southward) --> Na⁺(aq) + CH_iiiCOO^-(aq)

CH₃COO^-(aq) + H₂O(l) --> CH₃COOH(aq) + OH^-

The latter reaction proceeds forward only to a pocket-sized extent; the equilibrium constant M is very minor.  The equilibrium expression for this reaction can exist used to estimate the pH of the salt solution.  Since acetate functions equally a weak base, the equilibrium constant is given the label Thou_b.

Soluble salts that contain cations derived from weak bases class solutions that are acidic.  The cation is the conjugate acid of a weak base.  For example, the ammonium ion is the conjugate acrid of ammonia, a weak base.  Therefore, a soluble salt, such every bit ammonium chloride volition release ammonium ions into the solution, which a few of these will interact with h2o, forming ammonia and the hydronium ion.

NH_ivCl(s) --> NH₄ ⁺(aq) + Cl^-(aq)

NH₄ ⁺ + H_twoO(l) --> NH_iii(aq) + H₃O⁺(aq)

The latter reaction gain forward only to a modest extent, the equilibrium constant Yard is very small.  The equilibrium expresion for this reaction can be used to estimate the pH of the salt solution.  Since the ammonium ion functions as a weak acid, the equilibrium constant is given the label 1000_a.

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Relationship between Yard_a and K_b of Cohabit Acrid-Base Pairs

The relationship between Ka and Kb for whatever cohabit acid-base pairs is as follows:

(Thousand_a)(K_b) = Chiliad_west

Where K_a is the ionization constant of the acid form of the pair, Thousand_b is the ionization abiding for the base form of the pair, and K_west is the ionization constant for h2o.  This equation is used to discover either K_a or K_b when the other is known.

Case:  The K_a for acetic acid is 1.seven x 10^-5.  What is the value of Thou_b for the acetate ion?

(one.7 ten 10^-5)(Grand_b) = 1 x 10^-fourteen
K_b = v.9 x 10^-x

Case: The M_b for aniline is iii.8 x 10^-x.  What is the value of K_a for the anilonium ion?

(K_a)(three.viii x 10^-10) = 1 x 10^-xiv
Thousand_a = 2.6 x ten^-5

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Computing the pH of a Salt Solution

To calculate the pH of a salt solution i needs to know the concentration of the common salt solution, whether the salt is an acidic, basic, or neutral table salt, the equation for the interaction of the ion with the water, the equilibrium expression for this interaction and the K_a or G_b value.

Example: Calculate the pH of a 0.500 M solution of KCN.  K_a for HCN is 5.8 ten 10^-10.

• First, write the equation for the dissolving process, and examine each ion formed to determine whether the salt is an acidic, basic, or neutral salt.

KCN(s) --> K⁺(aq) + CN^-(aq)

K⁺ is a neutral ion and CN^- is a basic ion.  KCN is a basic salt.

• Second, write the equation for the reaction of the ion with h2o and the related equilibrium expression.

CN^-(aq) + H₂O(l) --> HCN(aq) + OH^-(aq)

Thousand_b = [HCN][OH ^- ]
[CN^-]

• 3rd, utilise the given G_a for HCN to discover the value of K_b for CN^-.

(five.8 10 10^-10)(K_b) = 1 x 10^-14
K_b = 1.7 ten 10^-5

• Make an "Water ice" nautical chart to aid in the solution.  Let "x" stand for the amount of CN^- that interacts with the water.

	CN-(aq)	HCN(aq)	OH-(aq)
Initial Concentration	0.500 M	0 M	0 M
Change in Concentration	- x	+ x	+ x
Equilibrium Concentration	0.500 M - x	x	x

• Subsititute equilibrium values and the value for Kb to solve for x.

1.seven x 10^-5 = (x)(10)/(0.500 - 10)

We will brand the supposition that since One thousand_b is so minor that the value for x will be very small also, thus the term (0.500 - 10) is equal to (0.500).

1.7 10 ten^-v = x²/(0.500)
x = 2.9 x 10^-iii

• Determine the pH of the solution.  Since "x" represents the hydroxide ion concentration, we can convert information technology into pOH and than find the pH.

pOH = -log(2.nine 10 10^-3) = 2.54
pH = fourteen - two.54 = xi.46

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Case:  What would be the pH of a 0.200 M ammonium chloride solution?  K_b ammonia = 1.viii x x^-5.

NH₄Cl(due south) --> NH₄ ⁺(aq) + Cl^-(aq)

NH_four ⁺ is an acidic ion and Cl^- is a neutral ion; solution volition exist acidic.

NH_iv ⁺(aq) + H_iiO(fifty) --> NH₃(aq) + H₃O⁺(aq)

K_a = [NH ₃ ][H ₃ O ⁺ ]
[NH₄ ⁺]

K_a = (1 ten 10^-xiv)/(1.8 x 10^-five) = five.6 x 10^-x

Let ten = the amount of NH_four ⁺ ion that reacts with the h2o.

	NHiv +(aq)	NHiii(aq)	H3O+(aq)
Initial Concentration	0.200 M	0	0
Change in Concentration	- ten	+ x	+ x
Equilibrium Concentration	0.200 K - x	x	x

5.half-dozen x 10^-10 = (10)(x)/(0.200 -ten)

5.6 x 10^-10 = x²/(0.200)

x = 1.1 ten ten^-5 M which is the H₃O⁺ concentration.

pH = -log(one.1 10 x^-v) = 4.98

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Case:  What is the pH of a 0.400 One thousand KBr solution?

KBr(s) --> One thousand⁺(aq) + Br^-(aq)

K⁺ and Br^- are both neutral ions.  The solution is neutral.

pH of the solution volition be 7.

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Source: https://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Salt_Solutions.htm

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