Sketch reasonably accurately the \(\mathrm{pH}\) curve for the titration of \(20.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) with \(0.20 \mathrm{M} \mathrm{KOH}(\mathrm{aq})\). Mark on the curve (a) the initial \(\mathrm{pH}\); (b) the pH at the stoichiometric point.

Short Answer

Expert verified
Initial pH is about 1 (from \(pH = -log(0.10)\)). At the stoichiometric point, the pH is 7. The curve starts at pH 1, rises sharply near the stoichiometric point, and levels off after reaching a pH above 7.

Step by step solution

01

Calculating Initial pH

Calculate the initial pH of the solution before any KOH is added. Since HCl is a strong acid, it will dissociate completely in solution. The pH is determined by the concentration of HCl. Use the formula \( pH = -\log[H^+] \) where \( [H^+] = 0.10 \, M \) (the molarity of HCl).
02

Identifying the Stoichiometric Point

The stoichiometric point is where the moles of HCl equal the moles of KOH. Calculate the volume of KOH needed to reach this point using the equation \( n_{HCl} = n_{KOH} \). This is equivalent to \( V_{HCl} \times M_{HCl} = V_{KOH} \times M_{KOH} \) and solve for \( V_{KOH} \) to obtain the volume of KOH needed to neutralize the acid.
03

Calculating pH at Stoichiometric Point

At the stoichiometric point, the solution is neutralized, resulting in a pH of 7 for the solution of a strong acid with a strong base.
04

Sketching the pH Curve

On a graph, sketch the pH curve starting from the initial pH calculated in Step 1. The pH will start low and begin to rise slowly as KOH is added. As the titration approaches the stoichiometric point, the curve will show a steep increase in pH. At and just after the stoichiometric point, mark the pH at 7. Continue the curve as it levels off at higher pH values after the equivalence point has been passed.
05

Annotating Initial and Stoichiometric pH

Label the starting point of the curve (a) with the initial pH. Then, label the point (b) where the pH equals 7, which is the stoichiometric point.

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Key Concepts

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

Acid-Base Titration
Acid-base titration is a methodology used to determine the concentration of an acid or a base by neutralizing it with an acid or base of known concentration. During this process, a solution, known as the titrant, is slowly added to another solution containing the substance being analyzed, termed the analyte.

Throughout the titration, the pH of the solution is monitored, usually using a pH meter or indicator solution, to find the point where the amount of titrant added precisely neutralizes the analyte. This is known as the equivalence point, and for a strong acid-strong base titration like our HCl-KOH example, it corresponds to a neutral pH of 7.

Understanding the Titration Curve

The curve produced in our exercise is a graphical representation showing the change in pH of the analyte solution as more titrant is added. Starting from the initial pH of the strong acid, the curve rises steeply nearby the equivalence point, reflecting the rapid change in pH as the solution transitions from acidic to neutral.
Stoichiometry
Stoichiometry is a branch of chemistry involved with the calculation of the quantitative relationships, or ratios, between the reactants and products in a chemical reaction. These ratios are derived from the balanced chemical equation for the reaction under study.

For the titration exercise, the reaction stoichiometry dictates that one mole of HCl reacts with one mole of KOH. The equation for this reaction is: \[\mathrm{HCl}(\mathrm{aq}) + \mathrm{KOH}(\mathrm{aq}) \rightarrow \mathrm{KCl}(\mathrm{aq}) + \mathrm{H_2O}(l)\]. By knowing the stoichiometry and the molar concentrations of the reactants, we can calculate essential values such as the volume of KOH needed to reach the stoichiometric point, where the moles of HCl have been exactly neutralized by the moles of KOH.

Applying Stoichiometry to Titration

By applying the stoichiometry of the neutralization reaction, we gain insights that allow us to determine the initial and final points on the titration curve and predict how the pH will change throughout the process.
Neutralization Reaction
A neutralization reaction is a specific type of chemical reaction where an acid and a base react to form water and a salt. In our example involving HCl and KOH, the acid (HCl) and the base (KOH) react to produce H2O and the salt KCl.

The concept of neutralization plays a pivotal role in understanding acid-base titrations. At the stoichiometric point of a titration curve, the moles of acid and base in the solution are equal, leading to the neutralization of both, reflected by a pH of 7 for strong acid-strong base reactions. This point is indicated on the curve and constitutes the moment when all the acid has been completely neutralized by the base.

The Result of Neutralization

After the equivalence point is reached in our exercise, adding more KOH will result in the mixture becoming progressively more basic. The titration curve will level off once again as excess base has less of an effect on the overall pH, indicative of the completion of the neutralization reaction.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) at each stage in the titration in which \(0.116 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) is added to \(25.0 \mathrm{~mL}\) of \(0.215 \mathrm{M} \mathrm{KOH}(\mathrm{aq})(\mathrm{a})\) initially; (b) after the addition of \(5.0 \mathrm{~mL}\) of acid; (c) after the addition of a further \(5.0 \mathrm{~mL}\); (d) at the stoichiometric point; (e) after the addition of \(5.0 \mathrm{~mL}\) of acid beyond the stoichiometric point; \((f)\) after the addition of \(10 \mathrm{~mL}\) of acid beyond the stoichiometric point.

A 0.968-g sample of impure sodium hydroxide was dissolved in \(200 \mathrm{ml}\). of aqueous solution. A \(20.0-\mathrm{mL}\). portion of this solution was titrated to the stoichiometric point with \(15.8 \mathrm{~mL}\) of \(0.107 \mathrm{M} \mathrm{HCl}\) (aq). What is the percentage purity of the original sample?

A 20-mL sample of \(0.020 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) was titrated with \(0.035 \mathrm{M} \mathrm{KOH}(\mathrm{aq})\). Calculate the \(\mathrm{pH}\) at the following points in the titration and sketch the \(\mathrm{pH}\) curve: (a) no \(\mathrm{KOH}\) added; (b) \(5.00 \mathrm{~mL}\) of \(\mathrm{KOH}\) (aq) added; (c) an additional \(5.00 \mathrm{~mL}\) of \(\mathrm{KOH}\) (aq) (for a total of \(10.0 \mathrm{~mL}\).) added; (d) another \(5.0 \mathrm{~mL}\) of \(\mathrm{KOH}(\mathrm{aq})\) added; (e) another \(5.00 \mathrm{~mL}\). \(\mathrm{KOH}(\mathrm{aq})\) added. (f) Determine the volume of \(\mathrm{KOH}\) (aq) required to reach the stoichiometric point.

Determine the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of (a) a solution that is \(0.40 \mathrm{M} \mathrm{NaHSO}_{4}(\mathrm{aq})\) and \(0.080 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\); (b) a solution that is \(0.40 \mathrm{M} \mathrm{NaHSO}_{4}(\mathrm{aq})\) and \(0.20 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq}) ;\) (c) a solution that is \(0.40 \mathrm{M} \mathrm{NaHSO}_{4}(\mathrm{aq})\) and \(0.40 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\).

Limestone is composed primarily of calcium carbonate, A \(1.0-\mathrm{mm}^{3}\) chip of limestone was accidentally dropped into a water-filled swimming pool, measuring \(10 \mathrm{~m} \times 7 \mathrm{~m} \times 2 \mathrm{~m}\). Assuming that the carbonate ion does not function as a Bronsted base and that the \(\mathrm{pH}\) of the water is 7 , will the pebble dissolve entirely? The density of calcium carbonate is \(2.71 \mathrm{~g} \cdot \mathrm{cm}^{-3}\).

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