Problem 48 - Chemistry: An Atoms-Focused Approach

Question

What volume of \(0.0100 M\) HCl is required to titrate \(250 \mathrm{mL}\) of \(0.0100 M \mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(250 \mathrm{mL}\) of \(0.0100 \mathrm{MHCO}_{3}^{-} ?\)

Step-by-Step Solution

Answer

Answer: 750 mL

Step 1: Calculate moles of Na2CO3 and HCO3-

To find the moles of \(Na_{2}CO_{3}\) and \(HCO_{3}^{-}\), we will use the formula: Moles = Molarity × Volume For \(Na_{2}CO_{3}\): Moles\(_{Na_{2}CO_{3}}\) = \(0.0100\ M\) × \(0.250\ L\) = \(0.00250\ mol\) For \(HCO_{3}^{-}\): Moles\(_{HCO_{3}^{-}}\) = \(0.0100\ M\) × \(0.250\ L\) = \(0.00250\ mol\)

Step 2: Determine reactions and stoichiometry between substances and HCl

Na2CO3 reacts with 2 moles of HCl: \(Na_{2}CO_{3}\) + \(2HCl \rightarrow 2NaCl\) + \(H_{2}O\) + \(CO_{2}\) HCO3- reacts with 1 mole of HCl: \(HCO_{3}^{-}\) + \(HCl \rightarrow H_{2}O\) + \(CO_{2}\) + \(Cl^{-}\) The stoichiometry between Na2CO3 and HCl is 1:2, and between HCO3- and HCl is 1:1.

Step 3: Calculate required moles of HCl

To find the required moles of HCl, we will use the stoichiometry from the balanced reactions: Moles\(_{HCl\,(for\,Na_{2}CO_{3})}\) = Moles\(_{Na_{2}CO_{3}}\) × \(\frac{2\ mol\ HCl}{1\ mol\ Na_{2}CO_{3}}\) = \(0.00250\ mol\) × 2 = \(0.00500\ mol\) Moles\(_{HCl\,(for\,HCO_{3}^{-})}\) = Moles\(_{HCO_{3}^{-}}\) × \(\frac{1\ mol\ HCl}{1\ mol\ HCO_{3}^{-}}\) = \(0.00250\ mol\) Total moles of HCl = Moles\(_{HCl\,(for\,Na_{2}CO_{3})}\) + Moles\(_{HCl\,(for\,HCO_{3}^{-})}\) = \(0.00500\ mol\) + \(0.00250\ mol\) = \(0.00750\ mol\)

Step 4: Calculate the required volume of HCl

Now, we will use the molarity of HCl and the required moles to find the volume: Volume = \(\frac{Moles}{Molarity}\) Volume\(_{HCl}\) = \(\frac{0.00750\ mol}{0.0100\ M}\) = \(0.750\ L\) (or \(750\ mL\)) So, \(750\ mL\) of \(0.0100\ M\ HCl\) is required to titrate both \(250\ mL\) of \(0.0100\ M\ Na_{2}CO_{3}\) and \(250\ mL\) of \(0.0100\ M\ HCO_{3}^{-}\).

Key Concepts

Molarity

Molarity is a way to express the concentration of a solution. It tells us how many moles of a substance are present in one liter of solution. It is important for calculating the amounts of reactants and products in a chemical reaction. The formula is simple:

• Molarity (M) = Moles of solute / Volume of solution in liters

In our original exercise, we see that both the hydrochloric acid (HCl) and the sodium carbonate solutions (\( \text{Na}_2\text{CO}_3 \) and \( \text{HCO}_3^-\)) have a molarity of 0.0100 M. This uniform concentration helps simplify the calculations. For instance, to find the number of moles in a solution, multiply the molarity by the volume of the solution in liters. For example:

• Moles of \( \text{Na}_2\text{CO}_3 \) = 0.0100 M × 0.250 L = 0.00250 mol
• Moles of \( \text{HCO}_3^- \) = 0.0100 M × 0.250 L = 0.00250 mol

Understanding molarity is crucial for determining how much of a reactant, such as HCl, is needed to complete a chemical reaction.

Stoichiometry

Stoichiometry is a term that describes the calculation of reactants and products in chemical reactions. It is based on the balanced chemical equation, which shows the exact ratios of molecules or moles that react and are produced. When you examine a balanced equation, stoichiometry helps you predict how much of each substance will be consumed or formed.
Let's take a look at the reactions presented in the solution:

• \( \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \)
• \( \text{HCO}_3^- + \text{HCl} \rightarrow \text{H}_2\text{O} + \text{CO}_2 + \text{Cl}^- \)

From these equations, we can determine the stoichiometric ratios. The first reaction shows a 1:2 ratio between \( \text{Na}_2\text{CO}_3 \) and \( \text{HCl} \), meaning one mole of \( \text{Na}_2\text{CO}_3 \) reacts with two moles of \( \text{HCl} \).
The second reaction indicates a 1:1 ratio between \( \text{HCO}_3^- \) and \( \text{HCl} \) — each mole of \( \text{HCO}_3^- \) reacts with one mole of \( \text{HCl} \). By using these ratios, we calculate the total moles of \( \text{HCl} \) needed for the reactions, helping us determine the volume required for titration.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In a titration, we typically have an acid-base reaction, which is a double replacement reaction. This reaction type is characterized by the exchange of ions between the reacting acid and base.
In our exercise, the primary reactions involve:

• \( \text{Na}_2\text{CO}_3 \) neutralizing \( \text{HCl} \) to form \( \text{NaCl} \), water, and carbon dioxide.
• \( \text{HCO}_3^- \) reacting with \( \text{HCl} \) to produce water, carbon dioxide, and chlorine ions.

These reactions highlight the fundamental concept of how acids react with bases to form salt and water, a classic outcome in titration.
Writing balanced equations for these reactions ensures we correctly apply stoichiometric principles and understand the proportions of products formed from given amounts of reactants.This understanding is vital in laboratory practices and industrial applications involving chemical processes.