Question

10. A 20.00 mL sample of 0.150 mol/L ammonia (NH3(aq)) is titrated to the equivalence point by 20.0 mL of a solution of 0.150 mol/L of the strong acid hydroiodic acid (HI (aq)).
a) Write a balanced equation for the titration reaction.
b) What is the pH of the ammonia solution before the titration begins?
c) What is the pH at the equivalence point?

Answer 1

Answer:

$\large \boxed{\rm a)\, NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \, NH_{4}^{+}(aq) +\text{I}^{-}(aq);\,b)\,11.22;\, c)\, 5.19}$

Explanation:

a) Balanced equation

The balanced chemical equation for the titration is

$\large \boxed{\rm NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \, NH_{4}^{+}(aq) +\text{I}^{-}(aq)}$

b) pH at start

For simplicity, let's use B as the symbol for NH₃.

The equation for the equilibrium is

$\rm B + H_{2}O \, \rightleftharpoons\,BH^{+} + OH^{-}$

(i) Calculate [OH]⁻

We can use an ICE table to do the calculation.

                      B + H₂O ⇌ BH⁺ + OH⁻

I/mol·L⁻¹:     0.150               0         0

C/mol·L⁻¹:       -x                 +x       +x

E/mol·L⁻¹:  0.150 - x            x          x

$K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.150 - x} = 1.8 \times 10^{-5}$

Check for negligibility:

$\dfrac{0.150 }{1.8 \times 10^{-5}} = 8300 > 400\\\\x \ll 0.150$

(ii) Solve for x

$\dfrac{x^{2}}{0.150} = 1.8 \times 10^{-5}\\\\x^{2} = 0.150 \times 1.8 \times 10^{-5}\\x^{2} = 2.7 \times 10^{-6}\\x = \sqrt{2.7 \times 10^{-6}}\\x = \text{[OH]}^{-} = 1.64 \times 10^{-3} \text{ mol/L}$

(iii) Calculate the pH

$\text{pOH} = -\log \text{[OH}^{-}] = -\log(1.64 \times 10^{-3}) = 2.78\\\\\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.78 = \mathbf{11.22}\\\\\text{The pH of the solution at equilibrium is } \large \boxed{\mathbf{11.22}}$

(c) pH at equivalence point

(i) Calculate the moles of each species

$\text{Moles of B} = \text{Moles of HI} = \text{20.00 mL} \times \dfrac{\text{0.0150 mmol}}{\text{1 mL}} = \text{3.00 mmol}$

                 B    +    HI   ⇌   BH⁺ + I⁻

I/mol:       3.00    3.00         0

C/mol:    -3.00   -3.00     +3.00

E/mol/:       0          0          3.00

(ii) Calculate the concentration of BH⁺

At the equivalence point we have a solution containing 3.00 mmol of NH₄I

Volume = 20.00 mL + 20.00 mL = 40.00 mL

$\rm [BH^{+}] = \dfrac{\text{3.00 mmol}}{\text{40.00 mL}} = \text{0.0750 mol/L}$

(iii) Calculate the concentration of hydronium ion

We can use an ICE table to organize the calculations.

                      BH⁺+ H₂O ⇌ H₃O⁺ +  B

I/mol·L⁻¹:     0.0750                 0        0

C/mol·L⁻¹:        -x                     +x      +x

E/mol·L⁻¹:   0.0750 - x             x         x

$K_{\text{a}} = \dfrac{K_{\text{w}}} {K_{\text{b}}} = \dfrac{1.00 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\\\\\dfrac{x^{2}}{0.0750 - x} = 5.56 \times 10^{10}\\\\\text{Check for negligibility of }x\\\dfrac{0.0750}{5.56 \times 10^{-10}} = 1.3 \times 10^{6} > 400\\\\\therefore x \text{ \ll 0.0750}$

$\dfrac{x^{2}}{0.0750} = 5.56 \times 10^{-10}\\\\x^{2} = 0.0750 \times 5.56 \times 10^{-10}\\x^{2} = 4.17 \times 10^{-11}\\x = \sqrt{4.17 \times 10^{-11}}\\\rm [H_{3}O^{+}] =x = 6.46 \times 10^{-6}\, mol \cdot L^{-1}$

(iv) Calculate the pH

$\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{6.46 \times 10^{-6}} = \large \boxed{\mathbf{5.19}}$

The titration curve below shows the pH at the beginning and at the equivalence point of the titration.