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Chapter 4: Problem 91

Explain why a titration experiment is a good way to measure the unknown concentration of a compound in solution.

Short Answer

Expert verified
A titration experiment is an effective method to measure the unknown concentration of a compound in a solution due to its principle of stoichiometry, precision in titrant addition, and endpoint detection. The stoichiometry of the reaction between the two solutions determines the ratio in which they react, allowing the calculation of the unknown concentration. The precision of titrant addition and effective endpoint detection ensures accurate and reliable results, making titration a suitable method for determining concentrations of unknown solutions.

Step by step solution

01

Understand Titration

Titration is a laboratory technique used to determine the concentration of an unknown solution by reacting it with a known volume and concentration of another solution, called the titrant. During a titration experiment, the titrant is slowly added to the unknown solution until the reaction is complete, as indicated by an endpoint. The endpoint is often observed using a pH indicator or by a sudden change in a chemical property of the solution.
02

Explain the Principle of Titration

In a titration experiment, the key principle is the stoichiometry of the reaction between the two solutions. The stoichiometric coefficients of the reactants determine the ratio in which they react and hence, can be used to calculate the unknown concentration of the compound in the solution.
03

Point out the Precision of Titration

Titration experiments are known for their accuracy and precision in measuring the unknown concentration of a compound in a solution. By utilizing a high degree of control in the volume of the titrant added to the unknown solution and a precise method to detect the endpoint, the concentration determined through titration has minimal room for error.
04

Explain the Endpoint Detection in Titration

One of the factors that make a titration experiment effective is the endpoint detection. A successful endpoint detection indicates that the equivalent point has been reached, which is the point where equal amounts of moles of the reactants have reacted in accordance with the stoichiometry principle. Several ways to detect the endpoint include the use of indicator dyes that change color at a specific pH or using electronic devices such as pH meters.
05

Conclude the Effectiveness of Titration in Measuring Concentration

Based on the explanation above, a titration experiment is a good way to measure the unknown concentration of a compound in a solution due to factors such as its principle of stoichiometry, precision in titrant addition, and the effectiveness in detecting the endpoint. These factors work together to enable the titration method to provide accurate and reliable results for determining the concentration of unknown solutions.

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Most popular questions from this chapter

You choose to investigate some of the solubility guidelines for two ions not listed in Table 4.1, the chromate ion \(\left(\mathrm{CrO}_{4}{ }^{2-}\right)\) and the oxalate ion \(\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right)\). You are given \(0.01 \mathrm{M}\) solutions (A, B, C, D) of four water-soluble salts: $$ \begin{array}{lll} \hline \text { Solution } & \text { Solute } & \text { Color of Solution } \\ \hline \mathrm{A} & \mathrm{Na}_{2} \mathrm{CrO}_{4} & \text { Yellow } \\ \mathrm{B} & \left(\mathrm{NH}_{4}\right)_{2} \mathrm{C}_{2} \mathrm{O}_{4} & \text { Colorless } \\ \mathrm{C} & \mathrm{AgNO}_{3} & \text { Colorless } \\ \mathrm{D} & \mathrm{CaCl}_{2} & \text { Colorless } \\ \hline \end{array} $$ When these solutions are mixed, the following observations are made: $$ \begin{array}{lll} \hline \text { Expt } & \text { Solutions } & \\ \text { Number } & \text { Mixed } & \text { Result } \\ \hline 1 & \mathrm{~A}+\mathrm{B} & \text { No precipitate, yellow solution } \\\ 2 & \mathrm{~A}+\mathrm{C} & \text { Red precipitate forms } \\ 3 & \mathrm{~A}+\mathrm{D} & \text { No precipitate, yellow solution } \\ 4 & \mathrm{~B}+\mathrm{C} & \text { White precipitate forms } \\ 5 & \mathrm{~B}+\mathrm{D} & \text { White precipitate forms } \\ 6 & \mathrm{C}+\mathrm{D} & \text { White precipitate forms } \\ \hline \end{array} $$ (a) Write a net ionic equation for the reaction that occurs in each of the experiments. (b) Identify the precipitate formed, if any, in each of the experiments. (c) Based on these limited observations, which ion tends to form the more soluble salts, chromate or oxalate?

Name the spectator ions in any reactions that may be involved when each of the following pairs of solutions are mixed. (a) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)\) and \(\mathrm{Mg} \mathrm{SO}_{4}(a q)\) (b) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) and \(\mathrm{Na}_{2} \mathrm{~S}(a q)\) (c) \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}(a q)\) and \(\mathrm{CaCl}_{2}(a q)\)

Separate samples of a solution of an unknown salt are treated with dilute solutions of \(\mathrm{HBr}, \mathrm{H}_{2} \mathrm{SO}_{4}\), and \(\mathrm{NaOH}\). A precipitate forms in all three cases. Which of the following cations could the solution contain: \(\mathrm{K}^{+} ; \mathrm{Pb}^{2+} ; \mathrm{Ba}^{2+}\) ?

(a) You have a stock solution of \(14.8 \mathrm{M} \mathrm{NH}_{3}\). How many milliliters of this solution should you dilute to make \(1000.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{NH}_{3} ?\) (b) If you take a \(10.0-\mathrm{mL}\) portion of the stock solution and dilute it to a total volume of \(0.500 \mathrm{~L}\), what will be the concentration of the final solution?

When asked what causes electrolyte solutions to conduct electricity, a student responds that it is due to the movement of electrons through the solution. Is the student correct? If not, what is the correct response?
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