Problem 31
Question
To a solution of acetic acid, solid sodium acetate is gradually added. When ' \(\mathrm{x} \mathrm{g}\) ' of the salt has been added, the \(\mathrm{pH}\) has a certain value. When total 'y g' of the salt has been added, the \(\mathrm{pH}\) has been further raised by \(0.6\) units. What is the ratio of \(x: y ?(\log 3.98=0.6)\) (a) \(3.98: 1\) (b) \(1: 3.98\) (c) \(2: 3.98\) (d) \(3.98: 2\)
Step-by-Step Solution
Verified Answer
The ratio of x to y is (a) 3.98:1.
1Step 1: Understand the Buffer System
Recognize that the solution of acetic acid and sodium acetate forms a buffer system. The pH change suggests that the acid-base equilibrium is being modified by the addition of the salt (sodium acetate).
2Step 2: Apply the Henderson-Hasselbalch Equation
Use the Henderson-Hasselbalch equation for buffer solutions: \[ pH = pKa + \log\left(\frac{[\text{A}^{-}]}{[\text{HA}]}, \right) \] where \( [\text{A}^{-}] \) is the concentration of acetate ion and \( [\text{HA}] \) is the concentration of acetic acid.
3Step 3: Calculate the Ratio from pH Change
Given that the pH increased by 0.6 units after adding 'y g' of the salt, we can set up two Henderson-Hasselbalch equations for the scenarios when 'x g' and 'y g' of the salt have been added, respectively, and find the ratio of acetate concentrations. Before the addition, the pH is: \[ pH = pKa + \log\left(\frac{x}{[\text{HA}]}, \right) \] After the addition, the pH is increased by 0.6: \[ pH + 0.6 = pKa + \log\left(\frac{y}{[\text{HA}]}, \right) \] Since we know \(\log 3.98 = 0.6\), we can infer that the ratio of the concentrations of acetate ions is 3.98:1. Therefore, the ratio of the masses of sodium acetate added, x to y, is also 3.98:1.
Key Concepts
pKa and pH Relationship
pKa and pH Relationship
The pKa value of a weak acid is a constant that represents its acid dissociation constant, describing its ability to donate a proton. The relationship between pH and pKa is paramount for understanding where the equilibrium lies in the solution.
When pH equals pKa, there is a 1:1 ratio of acid to conjugate base, meaning the solution is perfectly buffered against pH changes at this point. The farther away the pH is from the pKa, the less buffer capacity remains. In our exercise, since the buffer's pH increases as we add more sodium acetate, understanding the pKa of acetic acid allows us to predict just how much addition of the conjugate base (sodium acetate) will alter the pH. This understanding is crucial for procedures such as titrations, where precision in pH is necessary.
When pH equals pKa, there is a 1:1 ratio of acid to conjugate base, meaning the solution is perfectly buffered against pH changes at this point. The farther away the pH is from the pKa, the less buffer capacity remains. In our exercise, since the buffer's pH increases as we add more sodium acetate, understanding the pKa of acetic acid allows us to predict just how much addition of the conjugate base (sodium acetate) will alter the pH. This understanding is crucial for procedures such as titrations, where precision in pH is necessary.
Other exercises in this chapter
Problem 28
To \(20 \mathrm{ml}\) of \(0.1 \mathrm{M}-\mathrm{NaOH}\) solution, \(3 \mathrm{ml}\) of \(1 \mathrm{M}\) acetic acid solution is added. Is the solution now neu
View solution Problem 29
A volume of \(18 \mathrm{ml}\) of mixture of acetic acid and sodium acetate required \(6 \mathrm{ml}\) of \(0.1 \mathrm{M}-\mathrm{NaOH}\) for neutralization of
View solution Problem 32
Two buffers, \(X\) and \(Y\) of \(p H 4.0\) and \(6.0\) respectively are prepared from acid HA and the salt NaA. Both the buffers are \(0.50 \mathrm{M}\) in HA.
View solution Problem 32
The dissociation constant of formic acid is \(0.00024\). The hydrogen ion concentration in \(0.002 \mathrm{M}-\mathrm{HCOOH}\) solution is nearly (a) \(6.93 \ti
View solution