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.

Short Answer

Expert verified
pH at (a) is 1.7 (before any KOH added). As KOH is added, the pH increases progressively. The volume of KOH required to reach the stoichiometric point is 11.4 mL.

Step by step solution

01

Calculate the initial pH

To find the initial pH before any KOH is added, use the concentration of the HCl solution, which provides the concentration of hydronium ions, since HCl is a strong acid and dissociates completely in water. Use the formula pH = -log[H3O+]. In this case, pH = -log(0.020 M).
02

Calculate pH after 5.00 mL of KOH added

Calculate the moles of HCl and KOH, then find the moles of HCl remaining after the reaction. Use the new concentration of HCl to find the pH. Moles HCl initially = 20 mL * 0.020 M. Moles KOH added = 5.00 mL * 0.035 M. Moles HCl remaining = Initial moles HCl - moles KOH added. New concentration HCl = moles HCl remaining / total volume in L. pH = -log(new concentration of HCl).
03

Calculate pH after a total of 10.00 mL of KOH added

Repeat the previous steps considering the total volume of KOH added now is 10 mL. Moles KOH added = 10 mL * 0.035 M. Calculate moles of HCl remaining and pH as previously shown.
04

Calculate pH after an additional 5.00 mL of KOH added

Repeat the steps for additional 5 mL of KOH added, making the total added volume 15 mL, and calculate the moles of HCl and pH accordingly.
05

Calculate pH after another 5.00 mL of KOH added

Repeat the steps for another 5 mL of KOH added, making the total added volume 20 mL, and calculate the moles of HCl and pH accordingly.
06

Determine the stoichiometric point

The stoichiometric point is reached when moles of HCl = moles of KOH. Since both are 1:1 in stoichiometry, set moles HCl initially = moles KOH added and solve for the volume of KOH.

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Key Concepts

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

pH Calculation
Understanding how to calculate pH is fundamental in acid-base chemistry. pH stands for the 'potential of Hydrogen' or 'power of Hydrogen' and is a measure of the acidity or alkalinity of a solution. It is calculated on a logarithmic scale, where a lower pH value indicates a higher concentration of hydrogen ions (), making the solution more acidic. Conversely, a higher pH value signifies lower concentration and a more alkaline solution.

The formula to calculate pH is given by pH = -log(). For a strong acid like HCl, which dissociates completely in water, the pH can be directly calculated from its concentration. For example, if the concentration of HCl is 0.020 M, then the pH is -log(0.020 M). At each point in the titration with a base such as KOH, the pH can change depending on the amounts of acid and base present.
Stoichiometry
Stoichiometry is the section of chemistry that involves calculating the quantities of reactants and products in chemical reactions. When performing an acid-base titration, stoichiometry allows us to understand the mole relationship between the acid and the base involved. In a titration, the goal is often to determine the unknown concentration of a substance using a reaction for which the stoichiometry is known.

For the reaction between HCl and KOH, the stoichiometry is 1:1, meaning that one mole of HCl reacts with one mole of KOH. We use this information to help find the equivalence point of a titration, where the number of moles of acid equals the number of moles of base added. To calculate moles, we multiply the volume of the solution by its molarity. Stoichiometry is crucial for determining the amount of reactant remaining at different points during the titration, which in turn allows us to calculate pH.
Neutralization Reaction
A neutralization reaction is a type of chemical reaction in which an acid and a base react quantitatively with each other. In a simple acid-base neutralization, the protons from the acid combine with the hydroxide ions from the base to form water. This reaction can be represented by the general equation: Acid + Base → Salt + Water.

During an acid-base titration, the neutralization reaction occurs incrementally as the titrant is added to the analyte. In the case of HCl being titrated with KOH, the neutralization reaction produces KCl (a salt) and H2O. The process continues until all the available H+ ions from the acid have been consumed by OH- ions to form water, marking the end of the neutralization reaction.
Titration Curve
A titration curve is a graph that displays the change in pH as a function of the amount of titrant added during a titration. The curve typically starts with the initial pH of the solution being titrated and changes as the titrant is added.

In the titration of a strong acid with a strong base, the curve starts at a low pH due to the acidic solution. As base is added, the pH increases gradually until it reaches the equivalence point, where the amount of acid equals the amount of base added. At this point, the curve shows a steep rise in pH. Beyond the equivalence point, the solution becomes increasingly basic, and the pH rises more slowly. By analyzing the titration curve, one can determine the equivalence point and other significant data regarding the reaction.
Strong Acid and Base Titration
In a strong acid and base titration, both the acid and base fully dissociate in water, releasing their respective ions. When titrating a strong acid like HCl with a strong base such as KOH, the titration shows a clear and sharp equivalence point on the titration curve.

Since the stoichiometry is 1:1, the volume of KOH needed to reach the equivalence point can be found by equating the moles of KOH added to the initial moles of HCl. This type of titration allows for precise calculations of concentration and is frequently used in standardizing solutions and determining unknown concentrations. The clear end point, marked by a sharp pH change, is a characteristic feature of titrations involving strong acids and bases, making the analysis relatively straightforward.

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Most popular questions from this chapter

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 \(2.71 \mathrm{~g} \cdot \mathrm{cm}^{-3}\).

A 0.164-g sample of phosphorous acid, \(\mathrm{H}_{3} \mathrm{PO}_{3}\), is dissolved in water so that the total volume of the solution is \(50.0 \mathrm{~mL}\). (a) Estimate the \(\mathrm{pH}\) of this solution. (b) Fstimate the \(\mathrm{pH}\) of the solution that results when \(6.50 \mathrm{~mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH}(\mathrm{aq})\) is added to the phosphorous acid solution. (c) Fstimate the \(\mathrm{pH}\) of the solution if an additional \(4.93 \mathrm{~mL}\) of \(0.175 \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq})\) is added to the solution in part (b).

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 \(0.40 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\).

Galculate the volume of \(0.116 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) required to neutralize (a) one-half and (b) all the hydroxide ions in \(25.0 \mathrm{~mL}\) of \(0.215 \mathrm{M} \mathrm{KOH}\) (aq). (c) What is the molarity of \(\mathrm{Cl}^{-}\)ions at the stoichiometric point? (d) Calculate the \(\mathrm{pH}\) of the solution after the addition of \(40.0 \mathrm{~mL}\) of \(0.116 \mathrm{M}\) HCl(aq) to \(25.0 \mathrm{~mL}\) of \(0.215 \mathrm{M} \mathrm{KOH}(\mathrm{aq})\).

An old bottle labcled "Standardized \(6.0 \mathrm{M}\) \(\mathrm{NaOH}^{\prime \prime}\) was found on the back of a shelf in the stockroom. Over time, some of the NaOH had reacted with the glass and the solution was no longer \(6.0 \mathrm{M}\). To determine its purity, \(5.0 \mathrm{~mL}\) of the solution was diluted to \(100 \mathrm{ml}\). and titrated to the stoichiometric point with 11.8 mL. of \(2.05 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\). What is the molarity of the sodium hydroxide solution?

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