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Chapter 11: Problem 90

An old bottle labcled "Standardized \(6.0 \mathrm{M}\) \(\mathrm{NaOH}^{\prime \prime}\) was found on the back of a shelf in the stockroom. Over time, some of the NaOH had reacted with the glass and the solution was no longer \(6.0 \mathrm{M}\). To determine its purity, \(5.0 \mathrm{~mL}\) of the solution was diluted to \(100 \mathrm{ml}\). and titrated to the stoichiometric point with 11.8 mL. of \(2.05 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\). What is the molarity of the sodium hydroxide solution?

Short Answer

Expert verified
The molarity of the NaOH solution is the moles of NaOH (calculated in Step 4) divided by 0.005 L.

Step by step solution

01

Calculate moles of HCl used in titration

To find the molarity of the NaOH solution, start by calculating the moles of HCl used to reach the stoichiometric point. The molarity of the HCl is given as 2.05 M, and the volume used is 11.8 mL, which needs to be converted into liters for the calculation. Moles of HCl are found using the equation: moles = molarity × volume. Therefore, moles of HCl = 2.05 M × 11.8 mL × (1 L / 1000 mL).
02

Determine moles of NaOH

Knowing that the reaction between NaOH and HCl is a 1:1 stoichiometric reaction, where one mole of NaOH reacts with one mole of HCl, we can determine that the moles of NaOH originally in the 100 mL solution is equal to the moles of HCl used. So the moles of NaOH is also 2.05 M × 11.8 mL × (1 L / 1000 mL).
03

Calculate the original concentration of NaOH

To find the original molarity of NaOH, we need to calculate the concentration of the NaOH in the original 5.0 mL before dilution. Since the number of moles is constant, using the dilution formula: M1 × V1 = M2 × V2, where M1 is the original molarity, and V1 (5.0 mL) is the volume before dilution. Here, M2 is the molarity of NaOH in the titrated solution (unknown) and V2 (100 mL) is the volume after dilution. Using the moles of NaOH from Step 2, M1 can be calculated knowing that V1 is 5.0 mL (or 0.005 L).
04

Calculate the final Molarity of the NaOH solution

Now that we have the moles of NaOH and the volume before dilution, we can find M1, the original molarity of NaOH, using the equation: M1 = moles of NaOH / V1 (in liters). The final Molarity (M1) equals the moles of NaOH divided by 0.005 L.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molarity
Molarity, denoted by the symbol 'M', is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute dissolved in one liter of solution. The molarity equation is written as:
\[ M = \frac{moles\text{ of solute}}{volume\text{ of solution in liters}} \]
In the context of the exercise provided, we use molarity to describe the concentration of both the sodium hydroxide (NaOH) solution and the hydrochloric acid (HCl) solution. Understanding molarity is key to performing titration calculations as it allows us to find the relationship between the solute's quantity and the overall solution's volume.
Stoichiometry
Stoichiometry is a section of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is central to the makeup of a balanced chemical equation.
For the given exercise, the chemical reaction between NaOH and HCl is a stoichiometric one, meaning it proceeds in a 1:1 ratio: One mole of NaOH reacts with one mole of HCl.
By understanding this ratio, we can deduce that the moles of HCl used in the titration will be equal to the moles of NaOH present in the solution. This relationship is foundational to solving for the molarity of the NaOH solution.
Chemical Reaction
A chemical reaction is a process where reactants convert to products, often accompanied by a change in energy. In a titration, the chemical reaction of interest is typically an acid-base neutralization.
In our example, the neutralization reaction is between the base NaOH and the acid HCl to form water and sodium chloride (NaCl). Such reactions are typically characterized by an observable end point called the stoichiometric point, where the number of acid equivalents equals the number of base equivalents, indicating that the solution has been titrated to purity.
Dilution Formula
The dilution formula is used to calculate the concentration of a solution after it has been diluted. It is expressed as: \[ M1 \times V1 = M2 \times V2 \]
where M1 and V1 represent the molarity and volume of the original concentrated solution, and M2 and V2 represent the molarity and volume of the diluted solution, respectively. In the exercise, the dilution formula allows us to backtrack from the diluted solution's molarity to find the original molarity (concentration) of the NaOH solution prior to dilution.
This formula is pivotal in ensuring concentration consistency across a multitude of solution processes in chemistry.

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Most popular questions from this chapter

Determine the \(K_{\text {sp }}\) for the following sparingly soluble compounds, given their molar solubilities: (a) AgI, \(9.1 \times 10^{-9} \mathrm{~mol} \cdot \mathrm{L}^{-1}\); (b) \(\mathrm{Ca}(\mathrm{OH})_{2}\), \(0.011 \mathrm{~mol}-\mathrm{L}^{-1}\); (c) \(\mathrm{Ag}_{3} \mathrm{PO}_{4}, 2.7 \times 10^{-6} \mathrm{~mol}^{-\mathrm{L}}^{-1}\) (d) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}, 5.2 \times 10^{-7} \mathrm{~mol} \cdot \mathrm{L}^{-1}\).

A 20-mL sample of \(0.020 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) was titrated with \(0.035 \mathrm{M} \mathrm{KOH}(\mathrm{aq})\). Calculate the \(\mathrm{pH}\) at the following points in the titration and sketch the \(\mathrm{pH}\) curve: (a) no \(\mathrm{KOH}\) added; (b) \(5.00 \mathrm{~mL}\) of \(\mathrm{KOH}\) (aq) added; (c) an additional \(5.00 \mathrm{~mL}\) of \(\mathrm{KOH}\) (aq) (for a total of \(10.0 \mathrm{~mL}\).) added; (d) another \(5.0 \mathrm{~mL}\) of \(\mathrm{KOH}(\mathrm{aq})\) added; (e) another \(5.00 \mathrm{~mL}\). \(\mathrm{KOH}(\mathrm{aq})\) added. (f) Determine the volume of \(\mathrm{KOH}\) (aq) required to reach the stoichiometric point.

What volume of \(0.123 \mathrm{M} \mathrm{NaOH}(\mathrm{aq})\) must be added to \(125 \mathrm{~mL}\) of \(0.197 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{3}(\mathrm{aq})\) to reach (a) the first stoichiometric point; (b) the second stoichiometric point?

Which indicators could you use for a titration of \(0.20 \mathrm{M}\) acetic acid with \(0.20 \mathrm{M} \mathrm{NaOH}(\mathrm{aq})\) : (a) methyl orange; (b) litmus; (c) thymol blue; (d) phenolphthalein? Explain your selections.

A 0.164-g sample of phosphorous acid, \(\mathrm{H}_{3} \mathrm{PO}_{3}\), is dissolved in water so that the total volume of the solution is \(50.0 \mathrm{~mL}\). (a) Estimate the \(\mathrm{pH}\) of this solution. (b) Fstimate the \(\mathrm{pH}\) of the solution that results when \(6.50 \mathrm{~mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH}(\mathrm{aq})\) is added to the phosphorous acid solution. (c) Fstimate the \(\mathrm{pH}\) of the solution if an additional \(4.93 \mathrm{~mL}\) of \(0.175 \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq})\) is added to the solution in part (b).
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