Question

In an acid-base titration, \(25.62 \mathrm{mL}\) of an NaOH solution are needed to neutralize \(26.23 \mathrm{mL}\) of a \(0.1036 \mathrm{M} \mathrm{HCl}\) solution. To find the molarity of the NaOH solution, we can use the following procedure: a. First note the value of \(M_{\mathrm{H}^{+}}\) in the HCl solution. ________M b. Find \(M_{\mathrm{OH}^{-}}\) in the NaOH solution. (Use Eq. 3.) ________M c. Obtain \(M_{\mathrm{NaOH}}\) from \(M_{\mathrm{OH}^{-}}\) _______ M

Short answer

a. First note the value of \(M_{H^{+}}\) in the HCl solution. \(0.1036 \mathrm{M}\) b. Find \(M_{OH^{-}}\) in the NaOH solution. (Use Eq. 3.) \(0.1059 \mathrm{M}\) c. Obtain \(M_{\mathrm{NaOH}}\) from \(M_{\mathrm{OH}^{-}}\) \(0.1059 \mathrm{M}\)

Step-by-step solution

Note the Molarity of H+ ions in the HCl solution

The given molarity of HCl solution is 0.1036 M. Since HCl is a strong acid, it will dissociate completely, and the concentration of H+ ions will be the same as the concentration of the HCl solution. Therefore, the molarity of H+ ions, \(M_{H^+}\), is 0.1036 M.

Calculate the moles of H+ ions

We can now calculate the moles of H+ ions by using the formula: Moles of H+ ions = Molarity of H+ ions × Volume of HCl solution Moles of H+ ions = \(M_{H^+} \times V_{HCl}\) Moles of H+ ions = 0.1036 M × 0.02623 L (As 1 mL = 0.001 L, we convert 26.23 mL into liters by multiplying it by 0.001) Moles of H+ ions = 0.002715 mol

Calculate the moles of OH- ions

In a neutralization reaction, the moles of acid (H+) are equal to the moles of base (OH-). Therefore, the moles of OH- ions will be equal to 0.002715 mol.

Calculate the Molarity of OH- ions in the NaOH solution

Using the volume of the NaOH solution, we can find the molarity of OH- ions, by using the formula: Molarity of OH- ions = Moles of OH- ions / Volume of NaOH solution Molarity of OH- ions = \(M_{OH^{-}}\) \(M_{OH^{-}}\) = 0.002715 mol / 0.02562 L (As 1 mL = 0.001 L, we convert 25.62 mL into liters by multiplying it by 0.001) \(M_{OH^{-}}\) = 0.1059 M

Obtain the Molarity of the NaOH solution

Since NaOH is a strong base, it will dissociate completely, and the concentration of OH- ions will be the same as the concentration of the NaOH solution. Therefore, the molarity of NaOH, \(M_{NaOH}\), is equal to the molarity of OH- ions, which is 0.1059 M. So, the molarity of the NaOH solution is 0.1059 M.

Key concepts

Molarity Calculation

Molarity is a fundamental concept in chemistry often used to describe the concentration of a solution. It is represented as moles of solute per liter of solution. In the context of acid-base titration, molarity is crucial because it allows us to quantify the amount of acid or base present in a given volume of solution.

To calculate molarity, we use the formula: \[\begin{equation}Molarity (M) = \frac{Moles\ of\ solute}{Volume\ of\ solution\ in\ liters} \end{equation}\]For instance, in the provided exercise, we calculate the moles of the acid (HCl), and then we use the volume of the NaOH to find its molarity. It's important to convert the volume given in mL to liters before substituting in the formula, as molarity is defined in terms of liters. This foundation sets the stage for understanding the stoichiometry of the neutralization reaction.

Neutralization Reaction

A neutralization reaction is a type of chemical reaction where an acid and a base react with each other to form water and a salt. Specifically, the hydrogen ions (H+) from the acid combine with the hydroxide ions (OH-) from the base to produce water (H2O).

In the exercise, we see a classic neutralization reaction: \[\begin{equation}HCl + NaOH \rightarrow NaCl + H_2O\end{equation}\]Understanding that in a neutralization reaction, the moles of H+ ions will be equal to the moles of OH- ions, allows us to conclude that the number of moles of NaOH needed will be equivalent to HCl. This is due to the 1:1 molar ratio in their reaction, reflecting the principle that each H+ ion reacts with one OH- ion to produce water.

Stoichiometry

Stoichiometry is the section of chemistry that involves calculating the quantities of reactants and products in chemical reactions. In titration exercises like the one provided, we use stoichiometry to determine how much of one reactant (such as NaOH) is needed to completely react with a given amount of another reactant (such as HCl).

To solve problems involving stoichiometry, it's critical to understand the mole concept and be able to balance chemical equations. Once the equation is balanced, we can determine the molar relationships among the reactants and products and apply this information to ascertain the unknown molarity, as we did in the exercise with the NaOH solution.

Remember to verify that units match across the calculations—moles with moles, liters with liters—to ensure accurate results. The ability to expertly navigate stoichiometry is of immense value in not only titrations but nearly all areas of chemistry.