Question

Which of the following would AgBr have the greatest solubility? a.) 0.10 LiBr b.)0.10 M AgNO3 c.) 0.20 M NaBr d.) water

Calculate the pH of a 2.0x^-3 M solution HClO4 a.) 2.0 b.) 2.7 c.) 3.0 d.) 7.0

Which of the following are buffer systems: a.)NH4H2PO4/(NH4)2HPO4 b.) HCl/NaCl c.) H2SO4/Na2SO4 d.) HNO2/KNO2

Which of the following is the strongest weak acid? a.) NH4+ Ka=5.6x10^-10 b.) HCO3- Ka=5.6x10^-11 c.) H2CO3 Ka= 4.3x10^-7

What is true at equivalence point: a.) the maximum pH is reached b.) the minimum pH is reached c.) the stoichiometric relationship is utilized d.) none of the above

Please explain each answer!

Answer 1

(1) The solubility product, K_sp for dissociation of AgBr in a solution is given as

AgBr <=====> Ag⁺ + Br^- , K_sp = [Ag⁺][Br^-] (we ignore the concentration of solid AgBr as it is present in large excess).

From literature values, the K_sp for AgBr is 5.0*10^-13.

Now the molar solubility of Ag⁺:Br^-:AgBr is 1:1:1 (from the dissociation equation) which means that the solubility of the salt is same as that of the ions it dissociates into.

(a) In a 0.1 M LiBr solution, the concentration of Br^- from dissociation of AgBr is virtually low as compared to the Br^- produced from LiBr. Hence, [Br^- ] = 0.1 M.

Therefore,

5.0*10^-13 = [Ag⁺](0.1)

or, [Ag⁺] = 5.0*10^-12

The molar solubility of AgBr in a 0.1 M LiBr solution is 5.0*10^-12 M

(b) In a 0.10 M AgNO₃ solution, [Ag⁺] = 0.1 M. Solving, as above, we find the molar solubility of AgBr in 0.1 M AgNO₃ solution is 5.0*10^-12 M.

(c) In a 0.2 M NaBr solution, [Br^-] = 0.2 M; therefore,

5.0*10^-13 = [Ag⁺](0.2)

or, [Ag⁺] = 2.5*10^-12

The molar solubility of AgBr in a 0.2 M NaBr solution is 2.5*10^-12 M.

(d) In water, there are no common ions. Let the molar solubility of Ag⁺ =Br^- =x.

Therefore,

5.0*10^-13 = (x)(x)

or, x = 7.07*10^-7

The molar solubility of AgBr in water is 7.07*10^-7. Now since, higher the value of the solubility, the more soluble the compound is, hence, AgBr will have the greatest solubility in water.

(2) HClO₄ is a strong acid and is fully dissociated in water as

HClO₄ <======> H⁺ + ClO₄^-

Now, from the dissociation equation, we see that [HClO₄]:[H⁺]:[ClO₄^-] = 1:1:1, i.e, 1 mole of HClO₄ produces 1 mole of proton.

The molar concentration of the acid in our example is 2.0*10^-3 M.

Hence, [H⁺] = 2.0*10^-3

Therefore, pH = - log₁₀[H⁺] = -log₁₀(2.0*10^-3) = 2.698≈2.7

The pH of 2.0*10^-3 M HClO₄ is (b) 2.7

(3) A buffer system is formed between a weak acid and its conjugate base or a weak base and its conjugate acid.

(a) Both are ammonium salts, hence they cannot form a buffer system.

(b) HCl is a strong acid; hence it cannot form a buffer system.

(c) H₂SO₄ is a strong acid; hence it cannot form a buffer system.

(d) HNO₂ is a weak acid and KNO₂ is its conjugate base; hence they will form a buffer system.

(4) The higher the value of K_a of a weak acid, the greater is its tendency to donate protons. Since we say that an acid is more strong if it can donate more protons, hence of the three given weak acids, (c)H₂CO₃ (K_a = 4.3*10^-7) has the highest K_a value and hence it is the strongest weak acid.

(5) At the equivalence point, the moles of acid added are equivalent to the moles of base present or viceversa. Hence, (c) the stoichiometric relationship is utilized.