Consider the titration of 82.0 mL of 0.136 M Ba(OH)2 by 0.544 M HCl. Calculate the pH of the resulting solution after the following volumes of HCl have been added. (a) 0.0 mL WebAssign will check your answer for the correct number of significant figures. (b) 11.0 mL WebAssign will check your answer for the correct number of significant figures. (c) 32.0 mL WebAssign will check your answer for the correct number of significant figures. (d) 41.0 mL (e) 82.0 mL WebAssign will check your answer for the correct number of significant figures.

Explanation:

1. Write the overall equation.

The Grignard reaction has several steps:

(a) CH₃C₆H₄Br + Mg ⟶ CH₃C₆H₄MgBr

(b) CH₃C₆H₄MgBr + CO₂ ⟶ CH₃C₆H₄COOMgBr

(c) CH₃C₆H₄COOMgBr  + H⁺ ⟶ CH₃C₆H₄COOH + Mg²⁺ + Br⁻    

   CH₃C₆H₄Br + Mg + CO₂ + H⁺ ⟶ CH₃C₆H₄COOH + Mg²⁺ + Br⁻

Now we can gather all the information in one place.

Mᵣ:              171.03         24.31   44.01                        136.15

              CH₃C₆H₄Br +   Mg  +   CO₂ + H⁺ ⟶ CH₃C₆H₄COOH + Mg²⁺ + Br⁻

n/mol:        0.025

m/g:                                 0.7        10

Let's call CH₃C₆H₄Br "R" (reactant) and CH₃C₆H₄COOH "P" (product).

They have given us the amounts of three reactants and asked us to calculate the amount of product. This is a limiting reactant problem.

1. Calculate the moles of each reactant

(a) CH₃C₆H₄Br

n = 0.025 mol (given)

(b) Mg

(c) CO₂

2. Calculate the moles of CH₃C₆H₄COOH you can obtain from each reactant

The molar ratios are all 1:1, so

(a) 0.025 mol R    ⟶ 0.025 mol P

(b) 0.028 mol Mg ⟶ 0.029 mol P

(c) 0.23 mol CO₂  ⟶ 0.23    mol P

R is the limiting reactant, because it gives the fewest moles of P.

3. Calculate the theoretical yield.