The concentration of a certain sodium hydroxide solution was determined by using the solution to titrate a sample of potassium hydrogen phthalate (abbreviated as KHP). KHP is an acid with one acidic hydrogen and a molar mass of \(204.22 \mathrm{~g} / \mathrm{mol}\). In the titration, \(34.67 \mathrm{~mL}\) of the sodium hydroxide solution was required to react with \(0.1082\) g KHP. Calculate the molarity of the sodium hydroxide.

Short Answer

Expert verified
The concentration of the sodium hydroxide solution is approximately \(0.0153 \mathrm{M}\).

Step by step solution

01

Calculate moles of KHP

Using the given mass of KHP (\(0.1082 \mathrm{~g}\)) and its molar mass (\(204.22 \mathrm{~g/mol}\)), we can calculate the moles of KHP reacted in the titration. To do this, use the formula: moles of KHP = \(\frac{\text{mass of KHP}}{\text{molar mass of KHP}}\)
02

Find the moles of NaOH reacting with KHP

In the reaction, one mole of KHP reacts with one mole of sodium hydroxide (NaOH). Therefore, the moles of NaOH reacting with KHP will be the same as the moles of KHP. Moles of NaOH = Moles of KHP
03

Calculate the molarity of sodium hydroxide

Now that we have the moles of NaOH, we can calculate the molarity using the formula: Molarity = \(\frac{\text{moles of solute}}{\text{volume of solution in liters}}\) Since we have the volume of sodium hydroxide solution in milliliters (\(34.67 \mathrm{~mL}\)), we can convert it to liters by dividing by 1000. Now, let's put all the steps together and find the molarity of the sodium hydroxide solution.
04

Step 1

moles of KHP = \(\frac{0.1082 \mathrm{~g}}{204.22 \mathrm{~g/mol}} = 0.00053 \mathrm{~mol}\)
05

Step 2

Since Moles of NaOH = Moles of KHP: moles of NaOH = \(0.00053 \mathrm{~mol}\)
06

Step 3

First, convert the volume of sodium hydroxide solution to liters: Volume of NaOH solution = \(\frac{34.67 \mathrm{~mL}}{1000} = 0.03467 \mathrm{~L}\) Now, calculate the molarity of sodium hydroxide: Molarity of NaOH = \(\frac{0.00053 \mathrm{~mol}}{0.03467 \mathrm{~L}} = 0.0153 \mathrm{~M}\) Therefore, the concentration of the sodium hydroxide solution is approximately \(0.0153 \mathrm{M}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Molarity
Molarity is a vital concept in chemistry, representing the concentration of a solute in a solution. It is defined as the number of moles of a substance per liter of solution. When measuring molarity, the formula
\( M = \frac{\text{moles of solute}}{\text{volume of solution (L)}} \)
is used. This is a critical unit of concentration since it allows chemists to relate the volume of a solution to the amount of substance dissolved in it. Understanding molarity also helps determine how solutions will react when mixed, as seen in titration procedures. Molarity is especially useful because it provides a direct measure of the number of reacting particles in a given volume, which is essential for predicting the outcome of reactions.
Stoichiometry
Stoichiometry is the backbone of chemical reactions. It refers to the quantitative relationship between reactants and products in a chemical reaction. This aspect of chemistry enables scientists to predict the amounts of substances consumed and produced in a reaction, based on the balanced chemical equation. In a stoichiometric calculation, we use the coefficients of the balanced equation to understand the molar ratios of the reactants and products. For example, if a reaction indicates that one mole of substance A reacts with one mole of substance B, stoichiometry allows us to calculate how much of A is needed to fully react with a certain amount of B, or vice versa. It ensures the correct proportions are used for the reactants to avoid wasting materials and to predict the yield of the reaction.
Acid-Base Titration
Acid-base titration is a practical application of the concept of stoichiometry in analytical chemistry. In this process, a solution of known concentration (titrant) is gradually added to a solution of unknown concentration (analyte) until the chemical reaction between the two is complete – indicated by a color change or a pH meter reading.

Endpoint and Equivalence Point

An endpoint of a titration is when the indicator changes color, while the equivalence point is when the moles of acid equal the moles of base in the solution. These two points help chemists establish the point of complete neutralization, allowing for the calculation of the unknown concentration.
Titration calculations require precise measurement, a balanced reaction equation, and a thorough understanding of molarity and the stoichiometry of the reaction. By applying these principles, one can accurately determine the concentration of an unknown solution, vital in fields such as pharmaceuticals, environmental monitoring, and the food industry.
Molar Mass
Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). It is a fundamental property of substances that bridges the gap between the microscopic world of atoms and molecules and the macroscopic world that we can measure.
In chemistry, knowing the molar mass of a compound is crucial as it is used to convert between grams and moles, a key step in stoichiometry. The molar mass can be calculated by summing the masses of all the atoms in a molecule as given by the periodic table. For instance, to find the molar mass of a compound like KHP, we would add up the masses of potassium (K), hydrogen (H), and phthalate, a more complex group whose molar mass we would find by adding the masses of its constituent carbon, hydrogen, and oxygen atoms. An accurate molar mass allows for precise calculations in titrations and other quantitative analytical techniques.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CuSO}_{4}(a q)\) b. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CaCl}_{2}(a q)\) c. \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{MgI}_{2}(a q)\) d. \(\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q)+\mathrm{AlBr}_{3}(a q)\)

Saccharin \(\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S}\right)\) is sometimes dispensed in tablet form. Ten tablets with a total mass of \(0.5894 \mathrm{~g}\) were dissolved in water. The saccharin was oxidized to convert all the sulfur to sulfate ion, which was precipitated by adding an excess of barium chloride solution. The mass of \(\mathrm{BaSO}_{4}\) obtained was \(0.5032 \mathrm{~g}\). What is the average mass of saccharin per tablet? What is the average mass percent of saccharin in the tablets?

Douglasite is a mineral with the formula \(2 \mathrm{KCl} \cdot \mathrm{FeCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\). Calculate the mass percent of douglasite in a \(455.0-\mathrm{mg}\) sample if it took \(37.20 \mathrm{~mL}\) of a \(0.1000 \mathrm{M} \mathrm{AgNO}_{3}\) solution to precipitate all the \(\mathrm{Cl}^{-}\) as \(\mathrm{AgCl}\). Assume the douglasite is the only source of chloride ion.

Consider the reaction of \(19.0 \mathrm{~g}\) of zinc with excess silver nitrite to produce silver metal and zinc nitrite. The reaction is stopped before all the zinc metal has reacted and \(29.0 \mathrm{~g}\) of solid metal is present. Calculate the mass of each metal in the \(29.0-\mathrm{g}\) mixture.

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. \(0.0200 \mathrm{~mol}\) of sodium phosphate in \(10.0 \mathrm{~mL}\) of solution b. \(0.300 \mathrm{~mol}\) of barium nitrate in \(600.0 \mathrm{~mL}\) of solution c. \(1.00 \mathrm{~g}\) of potassium chloride in \(0.500 \mathrm{~L}\) of solution d. \(132 \mathrm{~g}\) of ammonium sulfate in \(1.50 \mathrm{~L}\) of solution

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free