Question

a. Determine the pH during the titration of 31.3 mL of 0.264 M trimethylamine ((CH₃)₃N, K_b = 6.3×10^-5) by 0.264 M HI at the following points. (Assume the titration is done at 25 °C.)
Note that state symbols are not shown for species in this problem.
(a) Before the addition of any HI ____

(b) After the addition of 12.6 mL of HI ____

(c) At the titration midpoint ____

(d) At the equivalence point ____

(e) After adding 43.8 mL of HI ____

b.

The K_sp for BaSO₄ is 1.1×10^-10 at 25 °C. Calculate the solubility of barium sulfate in pure water in (a) moles per liter and (b) grams per liter.

_____ moles/liter

_____ g/liter

Answer 1

Answer: a) Before addition of any HI, the following reaction can be written as B + H₂O --> BH⁺ + OH^-

B = (CH₃)₃N. The value of k_b for above reaction =  6.3×10^-5 (given)

From ICE table;

B ---> BH+ + OH-

Init 0.264M 0 0

change -x x x

Equil. (0.264-x) x x

k_b = [BH⁺][OH^-]/[B]

= x²/(0.264-x) or x² + k_b.x - 0.264k_b = 0 on solving quadratic equation we will get x = 0.004054

Now p(x) ~ pOH = -log(0.004054) =2.39

pH = 14 - pOH = 11.61

b. After the addition of 12.6 mL of HI

total volume = 43.9 mL (acid + base)

number of moles of acid present in 12.6 mL = 0.264/1000 x 12.6 mL = 3.3264 x 10^-3 moles (x₁)

number of moles of base present in 31.3 mL = 8.2632 x 10^-3 moles (x₂)

Following the above ICE table (OH^- will be used to neutralize HI);

B + H₂O   ----> BH⁺ + OH^-

Ini x₂     0 0

chan. -x₁     x₁ x₁

equi. x₂-x₁ 0 x₁   x1

Using henderson equation,

x₂-x₁ = 4.9368 x 10^-3 moles or [B] = x₂-x₁/0.0439 L = 0.1124 M

similarly [BH⁺] = x₁/0.0439 = 0.07577 M

pOH = pK_b + log [BH⁺]/[B] = 4.20 + log(0.6741) = 4.02 M

pH = 14.0 - pOH = 9.98

(c) At the titration midpoint ____
At midpoint [B] = [BH⁺] , Hence using the method involved in part b

pOH = pK_b = 4.2 as log term will be zero

pH = 9.8

(d) At the equivalence point ____

[B] = [HI] ; total volume = 31.3 (from B) + 31.3 (from HI) = 62.6 mL

The concentration of BH⁺ will determine the pH of solution.

So, the concentration of [BH⁺]  8.2632 x 10^-3 (moles of B in 31.3 mL of solution) / (0.0626 L) = 0.132 M

k_w = k_a.k_b from k_b for B, we can calculate the k_a for BH⁺ ; where k_w = 10^-14

k_a = 10^-14/6.3×10^-5 = 1.587 x10^-10  

Now the equilibrium reaction would be;

BH⁺ + H₂O --> B + H₃O⁺

ka = x.x/(0.132 - x) = x² +k_b.x - 0.132.k_b = 0

x = 4.5759 x 10^-6 ~ [H₃O⁺]

pH = -log[H₃O⁺] = 5.3395

e. After adding 43.8 mL of HI ____

The number of moles of HI present = 43.8 - 31.3 mL of acid (remaining part is consumed)

= 12.5 mL   

Moles of HI = 0.264/1000 x 12.5 = 3.3 x 10^-3 moles

Total volume = 31.3 + 43.8 = 75.1 mL

[HI] = 3.3 x 10^-3 /0.0751 L = 0.0439 M

For acid such as HI, assuming the dissociation to be ~100%

[H+] = 0.0439 M

pH = 1.35

Second Part The K_sp for BaSO₄ is 1.1×10^-10 at 25 °C

The K_sp relation for BaSO₄ = [Ba²⁺][SO₄^2-] = x.x = x².

x = sqrt(K_sp) = 1.0488 x 10^-5 M (moles/L)

Molar mass of BaSO₄ = 233.38 g/mol

Hence, strength of solution =  molarity x molar mass=  2.4476 x 10^-3 g/L

Please let me know if you have any doubt by commenting below the answer.

Thanks