/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 Potassium hydrogen phthalate, \(... [FREE SOLUTION] | 91Ó°ÊÓ

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Potassium hydrogen phthalate, \(\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4},\) is used to standardize solutions of bases. The acidic anion reacts with strong bases according to the following net ionic equation: $$ \mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow{\mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-}(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}(\ell) $$ If a \(0.902-\mathrm{g}\) sample of potassium hydrogen phthalate is dissolved in water and titrated to the equivalence point with \(26.45 \mathrm{mL}\) of \(\mathrm{NaOH},\) what is the molar concentration of the NaOH?

Short Answer

Expert verified
The molar concentration of NaOH is 0.167 M.

Step by step solution

01

Calculate Moles of Potassium Hydrogen Phthalate

First, we need to find the molar mass of potassium hydrogen phthalate (KHC₈H₄O₄). The atomic masses are approximately: K = 39.10, H = 1.01, C = 12.01, O = 16.00. Therefore, the molar mass is:\[ \text{Molar Mass of } \mathrm{KHC}_8\mathrm{H}_4\mathrm{O}_4 = 39.10 + 8(12.01) + 5(1.01) + 4(16.00) = 204.22 \text{ g/mol} \]Now, calculate the moles of the compound using the given mass:\[ \text{Moles of } \mathrm{KHC}_8\mathrm{H}_4\mathrm{O}_4 = \frac{0.902 \text{ g}}{204.22 \text{ g/mol}} = 0.004418 \text{ mol} \]
02

Use Stoichiometry of Reaction

In the given balanced net ionic equation, the stoichiometry between potassium hydrogen phthalate (as \( \mathrm{HC}_8\mathrm{H}_4\mathrm{O}_4^- \)) and hydroxide ions (\( \mathrm{OH}^- \)) is 1:1. This means the moles of \( \mathrm{OH}^- \) used in the reaction is equal to the moles of potassium hydrogen phthalate:\[ \text{Moles of } \mathrm{OH}^- = 0.004418 \text{ mol} \]
03

Calculate Molarity of NaOH

Molarity is defined as moles of solute divided by liters of solution. We have the moles of \( \mathrm{OH}^- \) and the volume of \( \mathrm{NaOH} \) solution used in the titration,\( 26.45 \text{ mL} = 0.02645 \text{ L} \). Use this to find the molarity:\[ \text{Molarity of NaOH} = \frac{0.004418 \text{ mol}}{0.02645 \text{ L}} = 0.167 \text{ M} \]

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Key Concepts

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

Potassium Hydrogen Phthalate
Potassium hydrogen phthalate (KHP), with the chemical formula \(\mathrm{KHC}_{8}\mathrm{H}_{4} \mathrm{O}_{4}\), is a crystalline, white solid commonly used in the laboratory as a primary standard for acid-base titrations. This compound contains a hydrogen phthalate ion \(\mathrm{HC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{-}\) that takes part in reactions with bases. Here are some key characteristics of this compound:
  • KHP is stable and non-hygroscopic, meaning it does not absorb moisture from the air, making it ideal for accurate mass measurements.
  • It has a high purity, which is crucial for being a reliable standard.
  • By dissolving a known mass of KHP in water, one can determine the concentration of basic solutions, such as NaOH, through titration.
In this exercise, \(0.902\text{ g}\) of KHP was dissolved in water to standardize a sodium hydroxide solution through titration.
Titration
Titration is a quantitative chemical analysis method used to determine the concentration of an unknown solution. The process involves gradually adding small volumes of a solution of known concentration, called the titrant, to the unknown solution until the chemical reaction is complete.
In this specific problem, sodium hydroxide (\(\mathrm{NaOH}\)) is used as the titrant to find its own concentration using potassium hydrogen phthalate (KHP) as the analyte.
  • The goal is to reach the equivalence point, the stage at which the amount of titrant is chemically equivalent to the amount of analyte in the solution.
  • Indicators, or pH meters, are often employed to detect this point.
The process in question reached the equivalence point when \(26.45\text{ mL}\) of \(\mathrm{NaOH}\) was added, indicating the consumed moles of hydroxide ions corresponded to those of KHP.
Net Ionic Equation
In chemical reactions, particularly in titrations, it's often helpful to simplify the equation to show only the species that actually change during the process. This simplification is captured by a net ionic equation. For the standardization of sodium hydroxide with potassium hydrogen phthalate, the net ionic equation is: \[ \mathrm{HC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{C}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{2-}(\mathrm{aq})+\mathrm{H}_{2}\mathrm{O}(\ell) \] This equation demonstrates that the hydrogen phthalate ion (\(\mathrm{HC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{-}\)) reacts with hydroxide ions (\(\mathrm{OH}^{-}\)) to form phthalate ions (\(\mathrm{C}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{2-}\)) and water.
Using net ionic equations helps focus on the actual chemical changes and can simplify stoichiometric calculations.
Molar Concentration Calculation
Determining the molarity or molar concentration of a solution is a critical skill in chemistry. The molarity is defined as the number of moles of solute divided by the volume of solution in liters.
In this exercise, we calculated the molarity of the \(\mathrm{NaOH}\) solution used in a titration process. Here's how it unfolds:
  • First, find out the moles of potassium hydrogen phthalate dissolved. Given its molar mass is \(204.22 \text{ g/mol}\), and \(0.902 \text{ g}\) is used, the moles of KHP are \(0.004418 \text{ mol}\).
  • The problem then uses stoichiometry to show that these moles of KHP equal the moles of \(\mathrm{OH}^{-}\) needed at the equivalence point.
  • With \(26.45\text{ mL}\) of \(\mathrm{NaOH}\) solution required, converted into liters (\(0.02645 \text{ L}\)), the molarity becomes \(\frac{0.004418 \text{ mol}}{0.02645 \text{ L}} = 0.167 \text{ M}\).
This calculation provides the molar concentration of the sodium hydroxide, essential for stoichiometric and analytical purposes.

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Most popular questions from this chapter

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