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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 g/ mol. In the titration, 34.67 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 molarity of the sodium hydroxide solution is 0.0153 M.

Step by step solution

01

Determine the moles of KHP

To determine the moles of KHP in the sample, we can use the following formula: moles of KHP = mass of KHP / molar mass of KHP We are given the mass of KHP (0.1082 g) and the molar mass of KHP (204.22 g/mol). moles of KHP = 0.1082 g / 204.22 g/mol Calculate the moles of KHP: moles of KHP = \( \frac{0.1082}{204.22} \) = 5.30 x 10^(-4) mol
02

Use stoichiometry to determine the moles of NaOH

From the balanced equation of the reaction between KHP and NaOH, we know that there's a 1:1 ratio of KHP to NaOH: KHP + NaOH → NaKP + H2O So, moles of NaOH = moles of KHP moles of NaOH = 5.30 x 10^(-4) mol
03

Calculate the molarity of the NaOH solution

Now, we will calculate the molarity of the NaOH solution using the moles of NaOH and the given volume of the solution that reacted with the KHP sample (34.67 mL). Remember, molarity (M) is calculated as: Molarity (M) = moles of solute / volume of solution (in liters) We need to convert the volume of NaOH in milliliters to liters before calculating the molarity: volume of NaOH (in L) = 34.67 mL / 1000 = 0.03467 L Now, we can calculate the molarity using the moles of NaOH and volume in liters: Molarity of NaOH = \( \frac{5.30 \times 10^{-4}\,\text{mol}}{0.03467\,\text{L}} \) = 0.0153 M The molarity of the sodium hydroxide solution is 0.0153 M.

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Key Concepts

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

Molarity
Molarity is a fundamental concept in chemistry that describes the concentration of a solute in a solution. It is expressed in terms of moles of solute per liter of solution, and is denoted by the symbol \( M \). Understanding molarity is crucial for various chemical applications, including titration. To calculate molarity, you follow this formula:
  • \( Molarity (M) = \frac{\text{moles of solute}}{\text{volume of solution (L)}} \)

In the context of a titration, molarity can determine the exact amount of reactant needed to neutralize a specific quantity of another substance. When you know the moles of solute and the volume of the solution, you calculate the molarity by simply dividing the two.
Consider our exercise where sodium hydroxide (NaOH) is used to titrate potassium hydrogen phthalate (KHP). By determining the molarity of NaOH, it helps us understand the concentration and hence the effectiveness of the solution in neutralizing the acid.
Sodium Hydroxide
Sodium hydroxide, often referred to by its chemical formula NaOH, is a strong base commonly used in chemistry labs for titration procedures. Its ability to react with acids to form water and a salt makes it particularly useful for neutralization experiments.
When used in titration, sodium hydroxide's concentration or molarity is a critical factor, as it determines how much of the solution is required to fully neutralize an acidic solution.
  • In our exercise, NaOH is used to determine the concentration of the solution by reacting with KHP, an acid.
  • The process involves gradually adding NaOH until all the acidic hydrogen from KHP is neutralized.

One crucial aspect to remember about sodium hydroxide is its hygroscopic nature—it can absorb moisture from the air, which might affect its molarity over time if not stored properly.
This property underscores the importance of precision and careful handling when preparing NaOH solutions for titration experiments.
Potassium Hydrogen Phthalate (KHP)
Potassium hydrogen phthalate, abbreviated as KHP, serves as a primary standard in acid-base titrations. Its chemical stability, high purity, and ease of weighing out precise quantities make it an ideal candidate for calibrating sodium hydroxide (NaOH) solutions.
KHP is a monoprotic acid, meaning it provides only one hydrogen ion (H+) per molecule for reaction. This simplifies calculations, as each mole of KHP reacts with one mole of NaOH in a straightforward 1:1 stoichiometric ratio.
  • In the original exercise, KHP's known molar mass of 204.22 g/mol is used to determine the amount in moles, which then reacts with NaOH.
  • This direct relation ensures that calculated moles of KHP directly indicate the moles of NaOH needed for neutralization.

Due to KHP's reliable behavior in titrations, it is favored for standardizing base solutions, ensuring accuracy in concentration measurements.

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Most popular questions from this chapter

You are given a solid that is a mixture of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) A 0.205-g sample of the mixture is dissolved in water. An excess of an aqueous solution of \(\mathrm{BaCl}_{2}\) is added. The BaSO\(_{4}\) that is formed is filtered, dried, and weighed. Its mass is 0.298 g. What mass of \(\mathrm{SO}_{4}^{2-}\) ion is in the sample? What is the mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample? What are the percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample?

Assign oxidation states for all atoms in each of the following compounds. a. \(\mathrm{UO}_{2}^{2+} \quad \quad f. \mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\) b. \(\mathrm{As}_{2} \mathrm{O}_{3} \quad \quad g. \mathrm{Na}_{2} \mathrm{P}_{2} \mathrm{O}_{3}\) c. \(\mathrm{NaBiO}_{3} \quad h. \mathrm{Hg}_{2} \mathrm{Cl}_{2}\) d. \(\mathrm{As}_{4} \quad\quad \quad i. \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) e. \(\mathrm{HAsO}_{2}\)

A stream flows at a rate of \(5.00 \times 10^{4}\) liters per second (L/s) upstream of a manufacturing plant. The plant discharges \(3.50 \times 10^{3} \mathrm{L} / \mathrm{s}\) of water that contains 65.0 \(\mathrm{ppm} \mathrm{HCl}\) into the stream. (See Exercise 135 for definitions.) a. Calculate the stream's total flow rate downstream from this plant. b. Calculate the concentration of \(\mathrm{HCl}\) in ppm downstream from this plant. c. Further downstream, another manufacturing plant diverts \(1.80 \times 10^{4} \mathrm{L} / \mathrm{s}\) of water from the stream for its own use. This plant must first neutralize the acid and does so by adding lime: $$\mathrm{CaO}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(i) $$ What mass of CaO is consumed in an 8.00-h work day by this plant? d. The original stream water contained 10.2 \(\mathrm{ppm} \mathrm{Ca}^{2+}\) . Although no calcium was in the waste water from the first plant, the waste water of the second plant contains \(\mathrm{Ca}^{2+}\) from the neutralization process. If 90.0% of the water used by the second plant is returned to the stream, calculate the concentration of \(\mathrm{Ca}^{2+}\) in ppm downstream of the second plant.

Specify which of the following are oxidation–reduction reactions, and identify the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced. a. \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \rightarrow 2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\) b. \(\mathrm{HCl}(g)+\mathrm{NH}_{3}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) c. \(\mathrm{SiCl}_{4}(i)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 4 \mathrm{HCl}(a q)+\mathrm{SiO}_{2}(s)\) d. \(\mathrm{SiCl}_{4}(l)+2 \mathrm{Mg}(s) \rightarrow 2 \mathrm{MgCl}_{2}(s)+\mathrm{Si}(s)\) e. \(\mathrm{Al}(\mathrm{OH})_{4}-(a q) \rightarrow \mathrm{AlO}_{2}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

What volume of each of the following acids will react completely with 50.00 mL of 0.200 M NaOH? a. 0.100 \({M} {HCl} \quad \)c. 0.200\(M {HC}_{2} {H}_{3} {O}_{2}$$(1 \text { acidic hydrogen })\) b. 0.150 \({M} {HNO}_{3}\)

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