Question

Base your answer on the information below and on your knowledge of chemistry.

A NaOH(aq) solution with a pH value of 13 is used to determine the molarity of an HCl(aq) solution. A 10.0-mL sample of the HCl(aq) is exactly neutralized by 16.0 mL of 0.100 M NaOH(aq). During this laboratory activity, appropriate safety equipment was used and safety procedures were followed.

Determine the pH value of a solution that has a H+(aq) ion concentration 10 times greater than the original NaOH(aq) solution.

Answer 1

Answer:

1. c = 0.16 mol·L⁻¹; 2. pH = 12

Explanation:

HCl + NaOH ⟶NaCl + H₂O

1. Concentration of HCl

(a)Moles of NaOH

$\text{Moles of NaOH} = \text{16.0 mL NaOH} \times \dfrac{\text{0.100 mmol NaOH}}{\text{1 mL NaOH}} = \text{1.6 mmol NaOH}$

(b) Moles of HCl

The molar ratio is 1 mol HCl:1 mol NaOH.  

$\text{Moles of HCl} = \text{1.6 mmol NaOH} \times \dfrac{\text{1 mmol HCl}}{\text{1 mmol NaOH}} = \text{1.6 mmol HCl}$

(c) Molar concentration of HCl

$c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{1.6 mmol }}{\text{10 mL}} = \textbf{0.16 mol/L}$

2. Hydronium ion in NaOH solution

In the original solution, pH = 13.

[H₃O⁺] = 10⁻¹³ mol·L⁻¹

At 10 times the concentration of hydronium ion

[H₃O⁺] = 10 × 10⁻¹³ mol·L⁻¹ = 10⁻¹² mol·L⁻¹

pH = -log[H₃O⁺] = -log(10⁻¹²) = 12

The pH decreases by one unit when the hydronium ion increases tenfold.