9.34, 9.38, 9.52, 9.56, 9.60 (a-b), 9.68, 9.74, 9.76

 9.34  An acid is classified as monoprotic, diprotic or triprotic based on whether it can dissociate to give one, two, or three hydrogen ions respectively in aqueous solution.

 (a) H₃PO₄, phosphoric acid, triporitc

(b) Ca(OH)₂ calcium hydroxide (base)

(c) HNO₃, nitric acid, monoprotic

(d) CH₃COOH, acetic acid, monoprotic

(e) KOH, potassium hydroxide (base)

(f) H₂CO₃, carbonic acid, diprotic

 9.38

acid = proton donor

base = proton acceptor 

Note: water can behave as either an acid or base

9.52

An equivalent is the mass of an acid (or base) that can produce one mole of H⁺ (or OH^-)

It is simply the molar mass of an acid (or base) divided by the number of H⁺ (or OH^-) each molecule of an acid (or base) can contribute. 

(a) LiOH – one mole can contribute one mole of OH^-: there is one equivalent per mole.  The molar mass of LiOH is 6.94+16.00+1.01=23.95 g / mol.

Hence, one equivalent of LiOH = 23.95g

(b) H₂SO₄ – one mole can contribute two moles of H⁺: there are two equivalents per mole.  The molar mass of H₂SO₄ is 2(1.01)+32.06+4(16.00) = 98.08 g / mol

Hence, one equivalent weight 49.04 g. 

(c) By similar reasoning, 29.2 g Mg(OH)₂

(d) By similar reasoning 32.7 g H₃PO₄

9.56

Remember, in a dilution calculations the number of moles of acid (or base) remains unchanged. 

Hence the moles initially present = M₁V₁ = moles after dilution = M₂V₂

The same holds true for the number of equivalents so

equivalents initially present = N₁V₁ = equivalents after dilution = N₂V₂ (N = normality)

N₁ = 2N, V₁ = ?

N₂ = 0.10N, V₂ = 250mL

V₁ = N₂V₂ / N₁ = (0.10N)(0.250L)/(2N) = 0.013L = 13mL

9.60

A titration is a neutralization reaction.  Enough acid (or base) of known normality is added to neutralize an acid (or base) of unknown molarity.  For neutralization, we must have the amount of OH^-  balance the amount of H⁺ or

equivalents OH^- = equivalents H⁺

As normality gives us equivalents, we have

V_baseN_base = V_acidN_acid

(a) V_acid = 15.0 mL = 0.0150 L, N_acid = 0.100N, V_base = 250 mL =  0.0250 L, N_base = ?

N_base = V_acidN_acid / V_base = (0.0150 L)(0.100 N) / (0.0250 L) = 0.060N

(b) V_base= 20.0mL = 0.0200L, N_base = 0.400N, V_acid = 20.0 mL = 0.0200 L

N_acid = V_baseN_base / V_acid = (0.0200 L)(0.400N) / (0.0200 L) = 0.400 N

9.68

A buffer system must be able to neutralize small quantities of strong acid and strong base by chemically reacting with H⁺ or OH^-.   A buffer consists of a weak acid and a weak base to provide a means of accomplishing this.  Basically, the addition of an acid or a base interconverts the weak acid and weak base that make up the buffer.  We can see that from the following two reactions:

Add base:

H₂PO₄^-(aq) + OH^-(aq) à HPO₄^2-(aq) + H₂O(l)

Add acid:

HPO₄^2-(aq) + H⁺(aq)  à H₂PO₄^-(aq) + H₂O (l)

9.74

Molarity is moles / Liter

Normaility is equivalents / Liter

We know the amount of H₂SO₃ and the volume of solution.   We just need to convert from grams to moles or equivalents. 

Molar mass of H₂SO₃ = 82.1 g / mol

9.76

To relate [H⁺] and [OH^-] we use the ion product constant of water

[H⁺][OH^-] = K_w = 1 x 10^-14

Calculate pH from [H⁺] by pH = -log[H⁺]

Calculate [H⁺] from pH by [H⁺] = antilog(-pH) = 10^-pH