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Consider a \(0.60-M\) solution of \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3},\) lactic acid \(\left(K_{\mathrm{a}}=\right.\) \(\left.1.4 \times 10^{-4}\right)\) a. Which of the following are major species in the solution? i. \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\) ii. \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{3}^{-}\) iii. \(\mathrm{H}^{+}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) \(\mathbf{v} . \mathrm{OH}^{-}\) b. Complete the following ICE table in terms of \(x,\) the amount (mol/L) of lactic acid that dissociates to reach equilibrium. c. What is the equilibrium concentration for \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{3}^{-} ?\) d. Calculate the \(p H\) of the solution.

Short Answer

Expert verified
The major species present in the solution are HC鈧僅鈧匫鈧, C鈧僅鈧匫鈧冣伝, H鈦, and H鈧侽. The equilibrium concentration of C鈧僅鈧匫鈧冣伝 is approximately 0.0093 M. The pH of the 0.60 M lactic acid solution is approximately 2.03.

Step by step solution

01

- Write down the dissociation equilibrium

Lactic acid, HC鈧僅鈧匫鈧,(a weak acid), will dissociate into the hydrogen ion, H鈦, and the lactate ion, C鈧僅鈧匫鈧冣伝. Write down the dissociation equilibrium as follows: \[HC_{3}H_{5}O_{3} \rightleftharpoons H^{+} + C_{3}H_{5}O_{3}^{-}\]
02

- Identify the major species in the solution

All species involved in dissociation of the lactic acid as well as water are considered major species in the solution. Thus, the major species present in the solution are i. HC鈧僅鈧匫鈧 ii. C鈧僅鈧匫鈧冣伝 iii. H鈦 iv. H鈧侽
03

- Set up the ICE table

Create an ICE table to track initial, change, and equilibrium concentrations of all species: | | HC鈧僅鈧匫鈧 | H鈦 | C鈧僅鈧匫鈧冣伝 | |---------------|-------------|------------|-----------| | Initial (M) | 0.60 | 0 | 0 | | Change (M) | -x | +x | +x | | Equilibrium(M)| 0.60 - x | x | x |
04

- Write the Ka expression

Use the Ka = 1.4 脳 10鈦烩伌 and the equilibrium concentrations from the ICE table to write the Ka expression: \[1.4 \times 10^{-4} = \frac{x \times x}{0.60 - x}\]
05

- Solve for x

Since Ka is small, we can approximate that x << 0.60, so 0.60 - x is nearly equal to 0.60. Hence, we can simplify and solve for x as follows: \[1.4 \times 10^{-4} = \frac{x^2}{0.60}\] \[x^2 = 1.4 \times 10^{-4} \times 0.60\] \[x = \sqrt{1.4 \times 10^{-4} \times 0.60}\] \[x \approx 0.0093 \,\text{M}\]
06

- Find the equilibrium concentration of C鈧僅鈧匫鈧冣伝

The equilibrium concentration of C鈧僅鈧匫鈧冣伝, from our ICE table, is equal to x: \[ [C_{3}H_{5}O_{3}^{-}]_{eq} = x \approx 0.0093\,\text{M}\]
07

- Calculate the pH of the solution

Since x is also the concentration of H鈦, calculate the pH using the formula pH = -log[H鈦篯: \[pH = -\log(0.0093) \approx 2.03\] So, the pH of the 0.60 M lactic acid solution is approximately 2.03.

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Key Concepts

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

Acid Dissociation Constant (Ka)
Understanding the acid dissociation constant, often abbreviated as Ka, is crucial when studying chemistry, particularly acid-base reactions. Ka measures the strength of an acid in solution, representing the equilibrium constant for the dissociation of an acid into its conjugate base and a proton (H+). A higher Ka value denotes a stronger acid, meaning that the acid donates protons more readily.

For lactic acid (HC鈧僅鈧匫鈧), the dissociation can be represented by the equation:
\[HC_{3}H_{5}O_{3} \rightleftharpoons H^{+} + C_{3}H_{5}O_{3}^{-}\]
Here, the Ka value is given as \(1.4 \times 10^{-4}\), which suggests that lactic acid is a weak acid, since its Ka value is relatively low, indicating it only partially dissociates in solution. When comparing to strong acids鈥攚ith Ka values much higher鈥攍actic acid produces fewer protons and its conjugate base in the solution at equilibrium.
ICE Table
The ICE table is an invaluable tool used to track the changes in concentrations of species in a chemical reaction, from the Initial state, through the Change, to the Equilibrium state. To illustrate this, let's use the dissociation of lactic acid.

Initially, we have a 0.60-M solution of lactic acid with no significant amount of H+ or lactate ions (C鈧僅鈧匫鈧冣伝). Upon dissociation, a certain amount \(x\) of lactic acid is converted into \(x\) moles per liter of H+ and \(x\) moles per liter of lactate ions. The ICE table helps us organize these thoughts:
  • Initial: Lactic acid at 0.60 M, H+ and lactate ions both at 0 M
  • Change: Lactic acid decreases by \(x\) M, H+ and lactate ions increase by \(x\) M
  • Equilibrium: Lactic acid at \(0.60 - x\) M, H+ and lactate ions both at \(x\) M

By setting up this framework, we can then apply our understanding of the acid dissociation constant to find the value of \(x\), which indicates how much lactic acid has dissociated at equilibrium.
pH Calculation
Calculating the pH of a solution is essential for understanding its acidity. The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration:\[pH = -\log[H^+]\]. In our example, \(x\) represents the equilibrium concentration of hydrogen ions produced by the dissociation of lactic acid.

To calculate the pH, we use the concentration of hydrogen ions that we obtained from solving the ICE table and insert it into the pH equation. For lactic acid, after approximating and simplifying the Ka expression, we found that \(x\) was approximately 0.0093 M, which is the concentration of hydrogen ions at equilibrium. Thus, the pH of the solution can be calculated:
\[pH = -\log(0.0093) \approx 2.03\]
As reflected by a pH of 2.03, the solution is acidic. This value is significant as it correlates with the weak acidic nature of lactic acid, which only partially dissociates in solution, releasing enough H+ to lower the pH below the neutral value of 7.

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Most popular questions from this chapter

Monochloroacetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2}\), is a skin irritant that is used in "chemical peels" intended to remove the top layer of dead skin from the face and ultimately improve the complexion. The value of \(K_{\mathrm{a}}\) for monochloroacetic acid is \(1.35 \times 10^{-3}\) Calculate the \(\mathrm{pH}\) of a \(0.10-M\) solution of monochloroacetic acid.

Calculate the \(\mathrm{pH}\) of each of the following solutions containing a strong acid in water. a. \(2.0 \times 10^{-2} M \mathrm{HNO}_{3}\) b. \(4.0 \mathrm{M} \mathrm{HNO}_{3}\) c. \(6.2 \times 10^{-12} \mathrm{M} \mathrm{HNO}_{3}\)

Rank the following 0.10 \(M\) solutions in order of increasing pH. a. \(\mathrm{NH}_{3}\) b. KOH c. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) d. KCl e. HCl

Calculate the \(\mathrm{pH}\) of a \(0.200-M\) solution of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\). Hint: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\) is a salt composed of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{F}^{-}\) ions. The principal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is $$\begin{aligned} \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \\ \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(a q) &+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\end{aligned}$$

Sodium azide (NaN, ) is sometimes added to water to kill bacteria. Calculate the concentration of all species in a \(0.010-M\) solution of \(\mathrm{NaN}_{3} .\) The \(K_{\mathrm{a}}\) value for hydrazoic acid \(\left(\mathrm{HN}_{3}\right)\) is \(1.9 \times 10^{-5}\)

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