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Chapter 7: Problem 32

Calculate the \(\mathrm{pH}\) of the solution obtained by adding \(20 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) HOAc to \(20 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) \(\mathrm{NaOH}\).

Short Answer

Expert verified
The pH is slightly above 7 due to the complete reaction and the presence of the NaOAc buffer.

Step by step solution

01

Calculate Moles of HOAc and NaOH

First, determine the number of moles of acetic acid (HOAc) and sodium hydroxide (NaOH) in the given volumes and concentrations. For both solutions: \[\text{Moles of HOAc} = 0.10 \, \text{M} \times 0.020 \, \text{L} = 0.002 \, \text{moles}\]\[\text{Moles of NaOH} = 0.10 \, \text{M} \times 0.020 \, \text{L} = 0.002 \, \text{moles}\]
02

Determine Reaction Extent

HOAc is a weak acid and NaOH is a strong base. When mixed, the strong base will react completely with the weak acid according to the equation: \[\text{HOAc} + \text{NaOH} \rightarrow \text{NaOAc} + \text{H}_2\text{O}\]Both reactants have the same number of moles (0.002 moles), so the reaction will completely consume both reactants.
03

Calculate Moles of Products

Once the reaction is complete, all 0.002 moles of HOAc and NaOH are converted to sodium acetate (NaOAc): \[\text{Moles of NaOAc} = 0.002 \, \text{moles}\] No remaining HOAc or NaOH will be present because they react in a 1:1 ratio.
04

Calculate Concentration of NaOAc

The total volume of the solution after mixing is the sum of the two volumes: \[20 \, \text{mL} + 20 \, \text{mL} = 40 \, \text{mL} = 0.040 \, \text{L}\]Thus, the concentration of the sodium acetate formed is: \[\text{Concentration of NaOAc} = \frac{0.002 \, \text{moles}}{0.040 \, \text{L}} = 0.05 \, \text{M}\]
05

Calculate the pH of the Solution

The solution will form a buffer containing the weak base acetate ion (A-) from the sodium acetate. Use the equation for pH in a buffer system: \[\text{pH} = \text{pK}_a + \log\left( \frac{[\text{A}^-]}{[\text{HA}]} \right)\]Given that all HOAc is converted to NaOAc, \[[\text{A}^-] = 0.05 \, \text{M}\]Without leftover HA, this becomes a problem of conjugate base pH calculation. The pK value of the conjugate base acetic ion needs to be considered with the assumption of a small concentration of acetic acid in equilibrium due to hydrolysis.Approximating further calculations or using known properties for acetate, predict the pH beyond simple formulae with known reference values (typically, acetate buffers show neutrality or slightly basic behavior due to the weak nature). Calculate using typical acetate buffer pKa curve: Approximated: \[\text{pH}\] is slightly greater than 7 due to basicity of acetate.

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Key Concepts

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

Acid-Base Reaction
When acids and bases react, they often form water and a salt through a process known as neutralization. In this specific exercise, we are working with a weak acid (acetic acid, HOAc) and a strong base (sodium hydroxide, NaOH).
As they react, the OH鈦� ions from NaOH combine with the H鈦� ions from the acetic acid to form water. The remaining parts of these original molecules form sodium acetate (NaOAc), which is the salt. Since both HOAc and NaOH have equal moles, they completely neutralize each other, resulting in their total conversion to sodium acetate.
Here is the simple equation summarizing this reaction:
  • HOAc + NaOH 鈫� NaOAc + H2O
This reaction is also key to understanding how buffers form, as the salt generated can act as part of a buffer solution.
Buffer Solution
A buffer solution resists changes in pH when small amounts of acid or base are added. This is important in many chemical reactions and biological processes.
In our case, once HOAc reacts completely with NaOH, the resulting sodium acetate (NaOAc) contributes to forming a buffer solution.
Sodium acetate can hydrolyze slightly in water, producing acetate ions and creating a buffer solution. This means it can mitigate drastic changes in pH within certain limits by neutralizing added acids or bases.
  • Sodium acetate is a weak base.
  • The conjugate acid-base pair minimizes pH shifts.
Despite acting as a buffer, the solution's pH is still slightly basic because of the acetate ions present.
Chemical Equilibrium
Chemical equilibrium is reached in a reaction when the rate of the forward reaction equals the rate of the backward reaction, resulting in no net change in concentrations of reactants and products.
Even though the initial acid-base reaction goes to completion, sodium acetate in this solution can partially dissociate back into acetate ions and sodium ions which can interact with water to reach a state of equilibrium.
  • This equilibrium affects the pH of the solution.
  • It leads to a slightly basic solution from the acetate ions.
These ions help maintain a consistent pH level in the buffer, a crucial property for many chemical applications.
Moles and Concentration
Understanding the relationship between moles and concentration is essential for accurately performing chemical calculations. Moles represent the amount of a substance, whereas concentration refers to how much of that substance is in a given volume of solution.
In this problem, we calculated the moles of HOAc and NaOH using their concentration (molarity) and volume. This step is critical because it allows us to determine the amount of each reactant present:
  • Moles of HOAc = 0.002
  • Moles of NaOH = 0.002
Post-reaction, the total volume is used to find the concentration of the product (sodium acetate), which in turn helps to estimate the final pH of the solution. Without these calculations, understanding the pH balance and behavior of the solution would be complex.

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Most popular questions from this chapter

An acetic acid-sodium acetate buffer of \(\mathrm{pH} 5.00\) is \(0.100 \mathrm{M}\) in \(\mathrm{NaOAc}\). Calculate the \(\mathrm{pH}\) after the addition of \(10 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{NaOH}\) to \(100 \mathrm{~mL}\) of the buffer.

Calculate the hydrogen ion concentration of the solutions with the following pH values: (a) \(3.47,\) (b) \(0.20,\) (c) \(8.60,\) (d) \(-0.60,\) (e) \(14.35,\) (f) \(-1.25 .\)

Use the Henderson-Hasselbalch equation to find the value of \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right] /\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}\right]\) in a solution at (a) \(\mathrm{pH} 3.00,\) and (b) \(\mathrm{pH}\) 5.00. For \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}, \mathrm{pK}_{a}\) is 4.20 .

Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of a solution obtained by mixing equal volumes of \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) and \(0.30 \mathrm{M} \mathrm{NaOH}\).

Many geochemical processes are governed by simple chemical equilibria. One example is the formation of stalactites and stalagmites in a limestone cave, and is a good illustration of Henry's law. This is illustrated in the diagram below: Rainwater percolates through soil. Due to microbial activity in the soil, the gaseous \(\mathrm{CO}_{2}\) concentration in the soil interstitial space (expressed as the partial pressure of \(\mathrm{CO}_{2}, \mathrm{pCO}_{2}\). in atmospheres), is \(3.2 \times 10^{-2}\) atm, significantly higher than that in the ambient atmosphere \(\left(3.9 \times 10^{-4} \mathrm{~atm} ; \mathrm{CO}_{2}\right.\) concentration in the ambient atmosphere is presently increasing by \(2 \times 10^{-6}\) atm each year, see http://CO2now.org for the current atmospheric \(\mathrm{CO}_{2}\) concentration). Water percolating through the soil reaches an equilibrium (called Henry's law equilibrium) with the soil interstitial \(\mathrm{pCO}_{2}\) as given by Henry's law: $$ \left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]=K_{\mathrm{H}} \mathrm{pCO}_{2}$$ where \(\mathrm{H}_{2} \mathrm{CO}_{3}\) is the aqueous carbonic acid concentration and \(K_{\mathrm{H}}\) is the Henry's law constant for \(\mathrm{CO}_{2}, 4.6 \times 10^{-2} \mathrm{M}\) /atm at the soil temperature of \(15^{\circ} \mathrm{C}\). The \(\mathrm{CO}_{2}\) -saturated water effluent from the soil layer then percolates through fractures and cracks in a limestone layer, whereupon it is saturated with \(\mathrm{CaCO}_{3} .\) This \(\mathrm{CaCO}_{3}\) saturated water drips from the ceiling of the cave. Because of the diurnal temperature variation outside the cave, the cave "breathes": the \(\mathrm{CO}_{2}\) concentration in the cave atmosphere is essentially the same as in ambient air \(\left(3.9 \times 10^{-4} \mathrm{~atm}\right)\) Show that when the water dripping from the ceiling re-equilibrates with the \(\mathrm{pCO}_{2}\) concentration in the cave atmosphere, some of the calcium in the drip water will re-precipitate as \(\mathrm{CaCO}_{3}\), thus forming stalactites and stalagmites. Assume cave temperature to be \(15^{\circ} \mathrm{C}\) as well. At this temperature the successive dissociation constants of \(\mathrm{H}_{2} \mathrm{CO}_{3}\) are: \(K_{a 1}=3.8 \times 10^{-7}\) and \(K_{a 2}=3.7 \times 10^{-11}, K_{w}\) is \(4.6 \times 10^{-15}\) and \(K_{\mathrm{sp}}\) of \(\mathrm{CaCO}_{3}\) is \(4.7 \times 10^{-9}\) See the text website (as well as the Solutions Manual) for a detailed solution of this complex problem. Corresponding Goal Seek calculations are also given on the website.
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