Problem 103 - Chemistry and Chemical Reactivity

Question

A solution of hydrochloric acid has a volume of \(125 \mathrm{mL}\) end a pH of \(2.56 .\) What mass of \(\mathrm{NaHCO}_{3}\) must be added to completely consume the HCl?

Step-by-Step Solution

Answer

Approximately 0.0289 grams of NaHCO₃ are needed.

Step 1: Calculate the Concentration of H⁺ Ions

The pH of a solution is given by the equation \[pH = -\log_{10}[H^+]\]To find the concentration of \([H^+]\), rearrange the equation:\[[H^+] = 10^{-pH} = 10^{-2.56} \]Now, calculate \([H^+]\): \[[H^+] \approx 2.75 \times 10^{-3} \, \text{M}\]

Step 2: Calculate the Moles of H⁺ Ions in the Solution

Use the concentration of \([H^+]\) to find the moles of \(H^+\) in the solution:\[moles = \text{Volume (L)} \times \text{Concentration (M)}\]Convert 125 mL to liters:\[125 \text{ mL} = 0.125 \text{ L}\]Calculate the moles of \(H^+\):\[\text{Moles of } H^+ = 0.125 \times 2.75 \times 10^{-3} = 3.44 \times 10^{-4} \text{ moles}\]

Step 3: Write the Chemical Reaction for Neutralization

The neutralization reaction between hydrochloric acid (HCl) and sodium bicarbonate (NaHCO₃) is:\[HCl + NaHCO_3 \rightarrow NaCl + CO_2 + H_2O\]This reaction involves a 1:1 molar ratio between \(HCl\) and \(NaHCO_3\).

Step 4: Calculate the Moles of NaHCO₃ Required

Since the reaction is a 1:1 molar ratio, the moles of \(NaHCO_3\) required will be the same as the moles of \(H^+\):\[\text{Moles of } NaHCO_3 = 3.44 \times 10^{-4} \text{ moles}\]

Step 5: Calculate the Mass of NaHCO₃ Needed

Find the mass of \(NaHCO_3\) required using its molar mass (84.01 g/mol):\[mass = \text{moles} \times \text{molar mass} = 3.44 \times 10^{-4} \times 84.01\]\[mass \approx 0.0289 \, \text{g}\]

Key Concepts

pH Calculation

Understanding the pH of a solution is crucial in chemistry, especially in acid-base chemistry. The pH scale measures the acidity or basicity of a solution. It ranges from 0 to 14, with 7 being neutral. Values less than 7 indicate acidity, and values greater than 7 indicate basicity.
In this exercise, the pH of hydrochloric acid (HCl) is given as 2.56. To find the concentration of hydrogen ions ([H^+]), you use the equation:\[pH = -\log_{10}[H^+]\]
Rearranging gives:\[[H^+] = 10^{-pH}\]
Calculating this for a pH of 2.56 results in a hydrogen ion concentration of approximately 2.75 \times 10^{-3} \, \text{M}\.
This tells us that the solution is quite acidic, as indicated by the low pH value and relatively high hydrogen ion concentration.

Moles and Stoichiometry

Converting measurements into moles is a fundamental skill in chemistry, often involving stoichiometry. Stoichiometry is the study of the relationships between the amounts of reactants and products in a chemical reaction. In this problem, knowing the moles of each component allows us to determine how much of a substance is needed for complete reaction.
Steps typically involve converting the volume from milliliters to liters and using the molarity to find moles, as done here:\[\text{Volume} = 125 \text{ mL} = 0.125 \text{ L}\]
Given the concentration [H^+] \approx 2.75 \times 10^{-3} \, \text{M}\, calculate the moles of H^+:\[\text{Moles of } H^+ = 0.125 \times 2.75 \times 10^{-3} = 3.44 \times 10^{-4} \text{ moles}\]
With stoichiometry, understanding the molar ratios between reactants is vital. In this exercise, the reaction between hydrochloric acid and sodium bicarbonate is a 1:1 molar ratio. Thus, you also need 0.000344 moles of sodium bicarbonate to neutralize the H^+ ions.

Acid-Base Chemistry

Acid-base reactions, known as neutralization reactions, involve acids and bases producing water and salt. These reactions are fundamental principles of chemistry.
The neutralization reaction, in this case, is described by:\[HCl + NaHCO_3 \rightarrow NaCl + CO_2 + H_2O\]
Here, hydrochloric acid (HCl), a strong acid, reacts with sodium bicarbonate (NaHCO_3), a weak base. This reaction forms sodium chloride (NaCl), water (H_2O), and carbon dioxide (CO_2).
The reaction highlights the 1:1 stoichiometric relationship between HCl and NaHCO_3. This means they react on a one-to-one basis, so the moles of HCl are equal to the moles of NaHCO_3 required for complete neutralization.
Calculating the mass of sodium bicarbonate involves:\[\text{mass} = \text{moles} \times \text{molar mass}\]
Given the molar mass of NaHCO_3 is 84.01 g/mol, multiply by the moles required:\[\text{Mass of } NaHCO_3 = 3.44 \times 10^{-4} \times 84.01\approx 0.0289 g\]
This calculation tells you exactly how much sodium bicarbonate is needed to neutralize the given amount of hydrochloric acid.