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A student carried out two titrations using an \(\mathrm{NaOH}\) solution of unknown concentration in the burette. In one titration, she weighed out \(0.2458 \mathrm{~g}\) of KHP ([see page \(166 .]\) ) and transferred it to an Erlenmeyer flask. She then added \(20.00 \mathrm{~mL}\) of distilled water to dissolve the acid. In the other titration, she weighed out \(0.2507 \mathrm{~g}\) of KHP but added \(40.00 \mathrm{~mL}\) of distilled water to dissolve the acid. Assuming no experimental error, would she obtain the same result for the concentration of the \(\mathrm{NaOH}\) solution?

Short Answer

Expert verified
Yes, she would obtain the same result for the concentration of NaOH.

Step by step solution

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01

Understanding the Reaction

KHP (potassium hydrogen phthalate) is a monoprotic acid, which means it will release one proton (H+) per molecule that reacts with NaOH. The balanced chemical equation is: \[ \text{KHP} + \text{NaOH} \rightarrow \text{NaKP} + \text{H}_2\text{O} \] This means that the moles of KHP will equal the moles of NaOH at the equivalence point.
02

Calculating Moles of KHP for Both Titrations

The molar mass of KHP is approximately \( 204.22 \, \text{g/mol} \). Therefore, the moles of KHP in the first titration are: \[ \text{Moles of KHP (first)} = \frac{0.2458 \, \text{g}}{204.22 \, \text{g/mol}} \approx 0.001203 \, \text{moles} \] For the second titration: \[ \text{Moles of KHP (second)} = \frac{0.2507 \, \text{g}}{204.22 \, \text{g/mol}} \approx 0.001228 \, \text{moles} \]
03

Concept of Titration and Volume of Water

During titration, the concentration of the NaOH is determined from the volume of NaOH needed to react completely with a known amount of KHP (in moles). The volume of distilled water used to dissolve KHP does not affect the moles of KHP or the volume of NaOH at the equivalence point. Therefore, it will not influence the calculated concentration of NaOH.
04

Conclusion of Titration Results

Since the moles of NaOH are determined by the moles of KHP and not by the volume of water used, both titrations will yield the same result for the concentration of the NaOH solution despite starting with different volumes of water.

Key Concepts

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

KHP (potassium hydrogen phthalate)
KHP, or potassium hydrogen phthalate, is a common acid used in titrations. It serves as a primary standard due to its stability and high purity. Its chemical formula is KHC₈H₄O₄, and it functions as a monoprotic acid. This means it donates one proton (H⁺) per molecule during a titration. When KHP is used in a titration, it reacts with a base (such as NaOH) in a one-to-one molar ratio. This predictable behavior makes it very useful for determining the concentration of the titrant, which in this case is NaOH. In your calculations, knowing the accurate mass of KHP and having its molar mass are crucial to finding the moles accurately.
Moles of NaOH
The moles of NaOH that react in a titration are directly related to the moles of KHP. Since we know the balanced chemical equation is KHP + NaOH → NaKP + H₂O, this means that every mole of KHP reacts with one mole of NaOH. To correctly determine the moles of NaOH, it's necessary to calculate the moles of KHP correctly. In this exercise, you find them by dividing the measured mass of KHP by its molar mass (204.22 g/mol). Therefore, knowing the moles of one reactant allows you to deduce the moles of the other.
Equivalence Point
The equivalence point in titration is when the quantity of titrant added is just sufficient to completely react with the analyte (the substance being measured). In the context of this exercise, it's the point at which all the KHP has reacted with NaOH. At the equivalence point, moles of KHP used wih the titrant are equal to the moles of NaOH added. This concept is key in titration as it indicates that the reaction has reached completion. Measuring the volume of NaOH solution used at the equivalence point allows for the determination of its concentration, provided you know the volume and the moles of KHP.
Molar Mass
Molar mass is a critical concept in chemistry that relates the mass of a substance to the amount of substance. To convert the grams of a substance into moles, you divide the mass by the molar mass. KHP has a molar mass of approximately 204.22 g/mol. Knowing this allows for accurate calculations of the moles of KHP from its mass. These moles are fundamental to finding out how much NaOH is needed to reach the equivalence point in titration. Thus, understanding molar mass helps bridge the gap between the measurable mass of a compound and the theoretical world of moles.

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Most popular questions from this chapter

Determine how many grams of each of the following solutes would be needed to make \(2.50 \times 10^{2} \mathrm{~mL}\) of a \(0.100-M\) solution: (a) cesium iodide (CsI), (b) sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right),(\mathrm{c})\) sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right),\) (d) potassium dichromate \(\left(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\right),\) (e) potassium permanganate \(\left(\mathrm{KMnO}_{A}\right)\)

Acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right)\) is a monoprotic acid commonly known as "aspirin." A typical aspirin tablet, however, contains only a small amount of the acid. In an experiment to determine its composition, an aspirin tablet was crushed and dissolved in water. It took \(12.25 \mathrm{~mL}\) of \(0.1466 \mathrm{M} \mathrm{NaOH}\) to neutralize the solution. Calculate the number of grains of aspirin in the tablet (one grain \(=0.0648 \mathrm{~g}\) ).

The current maximum level of fluoride that the EPA allows in U.S. drinking water is \(4 \mathrm{mg} / \mathrm{L}\). Convert this concentration to molarity.

Determine which of the following metals can react with acid: (a) \(\mathrm{Au},(\mathrm{b}) \mathrm{Ni},(\mathrm{c}) \mathrm{Zn},(\mathrm{d}) \mathrm{Ag},(\mathrm{e}) \mathrm{Pt}\).

For each of the following pairs of combinations, indicate which one will produce the greater mass of solid product: a) \(105.5 \mathrm{~mL} 1.508 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(250.0 \mathrm{~mL}\) \(1.2075 \mathrm{M} \mathrm{KCl}\) or \(138.5 \mathrm{~mL} 1.469 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(100.0 \mathrm{~mL} 2.115 \mathrm{M} \mathrm{KCl}\) b) \(32.25 \mathrm{~mL} 0.9475 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) and \(92.75 \mathrm{~mL} 0.7750 \mathrm{M}\) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) or \(52.50 \mathrm{~mL} 0.6810 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) and \(39.50 \mathrm{~mL} 1.555 \mathrm{M}\) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) c) \(29.75 \mathrm{~mL} 1.575 \mathrm{M} \mathrm{AgNO}_{3}\) and \(25.00 \mathrm{~mL} 2.010 \mathrm{M}\) \(\mathrm{BaCl}_{2}\) or \(52.80 \mathrm{~mL} 2.010 \mathrm{M} \mathrm{AgNO}_{3}\) and \(73.50 \mathrm{~mL} 0.7500 \mathrm{M}\) \(\mathrm{BaCl}_{2}\)

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