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Cacodylic acid is (CH3)2AsO2H. Its ionization constant is 6.4×107. (a) Calculate the pH of 50.0 mL of a 0.10M solution of the acid. (b) Calculate the pH of 25.0 mL of 0.15M(CH3)2AsO2Na. (c) Mix the solutions in parts (a) and (b). Calculate the pH of the resulting solution.

Short Answer

Expert verified
(a) pH = 3.10 (b) Calculate pH using conjugate base concentration. (c) Combine solutions and apply Henderson-Hasselbalch.

Step by step solution

01

Understand the Acid Dissociation in Part (a)

Cacodylic acid dissociates as (CH3)2AsO2H(CH3)2AsO2+H+. We have the initial concentration of the acid as [HA]=0.10M. Let's consider [H+]=x and make an ICE table to find x.
02

Set Up ICE Table for Cacodylic Acid

Initial: [HA]=0.10M,[A]=0,[H+]=0Change: [x,+x,+x]Equilibrium: [HA]=0.10x,[A]=x,[H+]=x\.
03

Calculate [H+] Using Ionization Constant

The ionization constant Ka=6.4×107. Using the equilibrium values, Ka=x20.10x. Assuming x0.10, Kax20.10.
04

Solve for x in Part (a)

Solve the equation 6.4×107=x20.10. Thus, x2=6.4×108 and x=8.0×104.
05

Calculate the pH for Part (a)

pH=log(8.0×104)=3.10.
06

Determine Concentration for Part (b)

(CH3)2AsO2 is the conjugate base present at 0.15M in 25 mL. This is a straightforward pH calculation using concentration of OH- ions.
07

Calculate pH for Part (b)

Since (CH3)2AsO2Na is a salt of a weak acid and strong base, it hydrolyzes to form (CH3)2AsO2 and pOH=log[OH], followed by pH=14pOH. Calculate [OH]=(Kw/Ka)0.15M.
08

Mix Both Solutions for Part (c)

Total moles of (CH3)2AsO2H from part (a) and total moles of (CH3)2AsO2 from (b) are determined. Then, find new concentrations in the combined solution volume.
09

Calculate pH of Final Mixture

Using concentrations from step above, apply Henderson-Hasselbalch equation: \(\text{pH} = \text{pK}_a + \log\left(\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]\right)\). Calculate pH.

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Key Concepts

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

Ionization Constant
The ionization constant, or acid dissociation constant (Ka), is a measure of the strength of an acid in solution. It provides insight into how well an acid molecule sheds its hydrogen ions, extH+, in a given reaction. A higher Ka value suggests a stronger acid that ionizes more completely in solution.

In the exercise involving cacodylic acid ( (CH3)2AsO2H), the ionization constant is given as 6.4×107. This implies that it is a weak acid since the value is far less than 1. Ka is used in equilibrium calculations to determine [H+], essential for later pH calculations. By setting up an equilibrium concentration expression, this small Ka can be used to find the rate at which the acid dissociates into its ions. This is crucial for predicting and calculating solution behaviors like acidity.
Acid Dissociation
Acid dissociation refers to the process by which an acid reacts with water to release hydrogen ions (H+) into the solution, forming its conjugate base. As seen in the example of cacodylic acid, (CH3)2AsO2H(CH3)2AsO2+H+.

Every acid dissociation involves an equilibrium between the non-dissociated acid molecule and the ions produced. The initial concentration of the acid, changes due to dissociation, and equilibrium concentrations can be calculated using an ICE table (Initial, Change, Equilibrium).
  • Initial: Start with the full concentration set as the acid form, for example, [HA]=0.10 M and zero for products.
  • Change: As dissociation occurs, the concentration of [HA] decreases by x, and [H+] and [A] increase by x.
  • Equilibrium: Calculate new concentrations using these changes, which allows for solving x, equating [H+] needed for pH.
Understanding this concept is essential as it forms the base for determining acid strength and eventual pH levels.
pH Calculation
The pH of a solution signals its acidity or basicity, mathematically defined as pH=log[H+]. For weak acids such as cacodylic acid, once [H+] is calculated from the Ka expression, pH can be directly derived from its concentration.

For example, using [H+]=8.0×104, pH=log(8.0×104)=3.10. This value is typical for weak acids, indicating a slightly acidic solution.

In addition, for basic solutions such as in the presence of a salt of a weak acid, calculate the [OH] instead, followed by using pOH=log[OH] and pH=14pOH. Such calculations are foundational for students to master the manipulation of logarithmic expressions essential in chemistry.
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch Equation is a pivotal tool in buffer solutions, formulated as:pH=pKa+log([A][HA])This formula directly links an acid's pKa with the ratio of its conjugate base ([A]) to the acid ([HA]) concentrations.

In the final exercise part where two solutions combine, creating a buffer, the Henderson-Hasselbalch equation becomes crucial. It bypasses more complex calculations to swiftly ascertain the final pH of a buffer solution. Here, knowing both initial concentrations and total solution volumes enables rapid pH calculation through this log-based equation.
  • pKa is determined from Ka as pKa=log(Ka).
  • Calculate the molar concentrations of [A] and [HA] in the new mixed solution volume.
  • Insert these values into the equation to find the resulting pH.
This method is immensely valuable when tackling buffer solutions, offering both accuracy and simplicity.

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