You are given \(250.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\) (propionic acid, \(\left.K_{\mathrm{a}}=1.35 \times 10^{-5}\right) .\) You want to adjust its \(\mathrm{pH}\) by adding an appropriate solution. What volume would you add of (a) \(1.00 \mathrm{M} \mathrm{HCl}\) to lower the \(\mathrm{pH}\) to \(1.00 ;\) (b) \(1.00 \mathrm{M} \mathrm{NaCH}_{3} \mathrm{CH}_{2} \mathrm{COO}\) to raise the \(\mathrm{pH}\) to \(4.00 ;(\mathrm{c})\) water to raise the \(\mathrm{pH}\) by 0.15 unit?

Short Answer

Expert verified
The volumes of the solutions required to adjust the pH are calculated based on the principles of equilibrium, stoichiometry, and acid-base chemistry. The volume of water needed to raise the pH by 0.15 was found using a dilution equation. Remember, this is a complex problem that will require time and effort. Do not get discouraged if it seems hard at first. Take it step by step, and keep practicing similar problems to improve your skills in solving them.

Step by step solution

01

Calculate initial pH

Firstly, calculate the initial pH of the propionic acid solution. This can be performed by setting up the pH equation, which is pH = -log[H+]. Since propionic acid is a weak acid, the [H+] or proton concentration can be obtained using the equation for the ionization constant (Ka) for weak acids. However, this concentration is same as the value of \( \sqrt {{K_a \cdot C}} \) where C is the concentration of the weak acid.
02

Determine volume of HCl for part (a)

The volume of 1.00 M HCl required to lower the pH to 1.00 can be calculated by first calculating the concentrations of HCl and propionic acid at pH = 1.00. This can be achieved using the pH equation. The proton concentration at this pH will be equal to 0.1 M (since pH 1 = -log[0.1]). Then, using the stoichiometry of the reaction, the volume of HCl (which is a strong acid and will dissociate fully) required can be calculated.
03

Determine volume of NaCH3CH2COO for part (b)

The amount of 1.00M NaCH3CH2COO necessary to raise the pH to 4.00 should be calculated next. To do this, we must first determine the [H+] at a pH of 4.00, which can be done using the logarithm. After that, the volume of Sodium propionate (NaCH3CH2COO) needed is calculated using a simplified form of the Henderson-Hasselbach equation, which is \( pH = pK_a + log(\frac {[Base]}{[Acid]}) \) where Base and Acid are the molarities of propanoic acid and sodium propanoate, respectively.
04

Calculate volume of water for part (c)

When it comes to calculating the amount of water required to raise the pH by 0.15 units, the calculations are a bit different. The increase in pH indicates a decrease in [H+]. So, we use the equation for dilution i.e, \( C1V1 = C2V2 \).\n Where C1= Initial [H+], C2= Final [H+], V1= Initial volume and V2= Final volume

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Key Concepts

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

Weak Acid Ionization
Understanding the behavior of weak acids is fundamental in chemistry, especially when calculating pH levels. Unlike strong acids that fully dissociate in water, weak acids do not completely ionize. This partial ionization is characterized by an equilibrium between the undissociated acid (HA) and its ions (H+ and A-). This process can be represented by an equilibrium constant known as the acid dissociation constant, Ka.
For example, with propionic acid (CH3CH2COOH), its ionization in water can be expressed as:
\[ CH3CH2COOH (l) + H2O (l) \rightleftharpoons CH3CH2COO^- (aq) + H^+ (aq) \]
The equilibrium constant for this reaction, Ka, can be calculated using the concentrations of the products and reactants at equilibrium. This constant is critical for calculating the initial pH of a weak acid solution and understanding how the pH will change when other substances are added.
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation establishes a relationship between pKa, pH, and the ratio of the concentration of an acid to its conjugate base. This equation greatly simplifies the process of calculating pH for buffer solutions, which consist of a weak acid and its conjugate base.
The equation is given as:
\[ pH = pKa + log \left( \frac{\text{[A^-]}}{\text{[HA]}} \right)\]
This relationship allows for the determination of the pH of a solution when the concentrations of an acid and its conjugate base are known or can be calculated. For instance, to raise the pH of a weak acid solution, one can add its conjugate base. The change in pH can be calculated by determining the new ratio of the base to the acid and then using the Henderson-Hasselbalch equation.
Acid-Base Stoichiometry
Acid-base stoichiometry involves the calculations based on the quantitative relationships between the amounts of reactants and products in an acid-base reaction. Knowing the molarity and volume of acidic and basic solutions allows for precise calculations of how much of one solution is needed to react completely with another.
When dealing with strong acids like HCl, the stoichiometry is straightforward since they fully dissociate into H+ and Cl-. To find the volume of HCl needed to lower the pH of a weak acid solution to a certain level, we apply the stoichiometry of the neutralization reaction:
\[ HA + HCl \to A^- + H_2O \]
Through this stoichiometric relationship, we calculate how much strong acid is required to achieve the desired pH, taking into account the concentration of hydrogen ions at that pH level.
pH and pKa Relationship
The pH of a solution is a measure of its acidity or alkalinity, expressed as the negative logarithm of the hydrogen ion concentration. For any acid, pKa is the negative logarithm of the acid dissociation constant, Ka. The relationship between pH and pKa is important for understanding how a solution's pH changes with the addition of acid or base.
A solution where the pH is equal to the pKa of the weak acid is where the concentrations of the acid and its conjugate base are equal. This point represents the maximum buffering capacity of the solution. Knowing this relationship helps in designing buffer solutions and predicting how the buffer will respond to the addition of acids or bases. For instance, adjusting the pH to a certain level involves adding substances that will shift the ratio of the weak acid to its conjugate base, thereby changing the pH according to the Henderson-Hasselbalch equation.
Molarity and Volume Relationship
Molarity (M) is the number of moles of solute per liter of solution, and it plays a crucial role in the dilution and concentration of solutions. The concept of molarity is essential when performing dilutions, which involve adding solvent to a solution to decrease the concentration of solutes.
The molarity and volume relationship is captured in the dilution equation:
\[ M_1V_1 = M_2V_2 \]
where \(M_1\) and \(M_2\) represent the initial and final molarities, respectively, while \(V_1\) and \(V_2\) represent the initial and final volumes. This equation is particularly useful when calculating how much water to add to a solution to change its pH, as in the case of raising the pH of the propionic acid solution by a small amount. By applying this dilution principle, one can determine the final volume of the solution needed to achieve the targeted molarity, and as a result, the desired pH.

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Most popular questions from this chapter

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) is a somewhat stronger acid than water. Its ionization is represented by the equation \(\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HO}_{2}^{-}\) In \(1912,\) the following experiments were performed to obtain an approximate value of \(\mathrm{p} K_{\mathrm{a}}\) for this ionization at \(0^{\circ} \mathrm{C} .\) A sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was shaken together with a mixture of water and 1 -pentanol. The mixture settled into two layers. At equilibrium, the hydrogen peroxide had distributed itself between the two layers such that the water layer contained 6.78 times as much \(\mathrm{H}_{2} \mathrm{O}_{2}\) as the 1 -pentanol layer. In a second experiment, a sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was shaken together with 0.250 M NaOH(aq) and 1-pentanol. At equilibrium, the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was \(0.00357 \mathrm{M}\) in the 1-pentanol layer and 0.259 M in the aqueous layer. In a third experiment, a sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was brought to equilibrium with a mixture of 1 -pentanol and \(0.125 \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq}) ;\) the concentrations of the hydrogen peroxide were \(0.00198 \mathrm{M}\) in the 1 -pentanol and \(0.123 \mathrm{M}\) in the aqueous layer. For water at \(0^{\circ} \mathrm{C}, \mathrm{p} K_{\mathrm{w}}=14.94\) Find an approximate value of \(\mathrm{pK}_{\mathrm{a}}\) for \(\mathrm{H}_{2} \mathrm{O}_{2}\) at \(0^{\circ} \mathrm{C}\) [Hint: The hydrogen peroxide concentration in the aqueous layers is the total concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) and \(\mathrm{HO}_{2}^{-}\). Assume that the 1 -pentanol solutions contain no ionic species.

The ionization constants of ortho-phthalic acid are \(K_{\mathrm{a}_{1}}=1.1 \times 10^{-3}\) and \(K_{\mathrm{a}_{2}}=3.9 \times 10^{-6}\) 1. \(\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{COOH})_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}\) 2\. \(\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{COO}^{-}\right)_{2}\) What are the pH values of the following aqueous solutions: (a) 0.350 M potassium hydrogen orthophthalate; (b) a solution containing 36.35 g potassium ortho-phthalate per liter? (f) \(0.68 \mathrm{M} \mathrm{KCl}, 0.42 \mathrm{M} \mathrm{KNO}_{3}, 1.2 \mathrm{M} \mathrm{NaCl},\) and \(0.55 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO},\) with \(\mathrm{pH}=6.4\)

Calculate the \(\mathrm{pH}\) of a \(0.5 \mathrm{M}\) solution of \(\mathrm{Ca}(\mathrm{HSe})_{2}\), given that \(\mathrm{H}_{2}\) Se has \(K_{\mathrm{a}_{1}}=1.3 \times 10^{-4}\) and \(K_{\mathrm{a}_{2}}=1 \times 10^{-11}\)

Solution (a) is \(100.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) and solution (b) is \(150.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{NaCH}_{3} \mathrm{COO}\). A few drops of thymol blue indicator are added to each solution. What is the color of each solution? What is the color of the solution obtained when these two solutions are mixed?

The neutralization of \(\mathrm{NaOH}\) by \(\mathrm{HCl}\) is represented in equation (1), and the neutralization of \(\mathrm{NH}_{3}\) by HCl in equation (2). 1. \(\mathrm{OH}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O} \quad K=?\) 2\. \(\mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{O}^{+} \rightleftharpoons \mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O} \quad K=?\) (a) Determine the equilibrium constant \(K\) for each reaction. (b) Explain why each neutralization reaction can be considered to go to completion.

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