Problem 27 - Chemistry: The Central Science

Question

A buffer solution contains \(0.10 \mathrm{~mol}\) of acetic acid and $0.13 \mathrm{~mol}\( of sodium acetate in \)1.00 \mathrm{~L}$. (a) What is the \(\mathrm{pH}\) of this buffer? (b) What is the \(\mathrm{pH}\) of the buffer after the addition of \(0.02\) mol of KOH? (c) What is the pH of the buffer after the addition of \(0.02 \mathrm{~mol}\) of \(\mathrm{HNO}_{3}\) ?

Step-by-Step Solution

Answer

(a) The initial pH of the buffer solution is calculated using the Henderson-Hasselbalch equation, resulting in: \[pH = 4.75 + \log{\frac{0.13}{0.10}} = 4.97\] (b) After adding \(0.02~mol\) of KOH, the pH of the buffer solution is: \[pH = 4.75 + \log{\frac{0.15}{0.08}} = 5.30\] (c) After adding \(0.02~mol\) of HNO3, the pH of the buffer solution is: \[pH = 4.75 + \log{\frac{0.13}{0.10}} = 4.97\]

Step 1: Calculate the initial pH of the buffer solution

To calculate the pH of the buffer, we'll use the Henderson-Hasselbalch equation: \[pH = pK_a + \log{\frac{[A^-]}{[HA]}}\] Here, \(pK_a\) represents the negative logarithm of the acid dissociation constant of acetic acid (\(K_a\)), which is \(4.75\). \([A^-]\) and \([HA]\) are the concentrations of the conjugate base (acetate ion, coming from sodium acetate) and the weak acid (acetic acid), respectively. By substituting the given values, we get: \[pH = 4.75 + \log{\frac{0.13}{0.10}}\] Calculate the value within the logarithm function and then add it to the \(pK_a\).

Step 2: Calculate the pH after adding KOH

First, we need to find the amount of acetic acid and acetate ion after the reaction with KOH. The equation for the reaction between acetic acid and KOH is: \(CH_3COOH + KOH \rightarrow CH_3COOK + H_2O\) KOH reacts completely with acetic acid as it is a strong base. Thus, \(0.02~mol\) of acetic acid will react with \(0.02~mol\) of KOH, producing \(0.02~mol\) of the acetate ion. Update the molar amounts of acetic acid and acetate ion: Moles of acetic acid: \(0.10 - 0.02 = 0.08~mol\) Moles of acetate ion: \(0.13 + 0.02 = 0.15~mol\) Now, calculate the pH using the Henderson-Hasselbalch equation: \[pH = 4.75 + \log{\frac{0.15}{0.08}}\] Calculate the value within the logarithm function and then add it to the \(pK_a\).

Step 3: Calculate the pH after adding HNO3

Now, we need to find the amount of acetic acid and acetate ion after the reaction with HNO3. The equation for the reaction between the acetate ion and HNO3 is: \(CH_3COO^- + HNO_3 \rightarrow CH_3COOH + NO_3^-\) Since HNO3 is a strong acid, it reacts completely with acetate ions. Thus, \(0.02~mol\) of the acetate ion will react with \(0.02~mol\) of HNO3, producing \(0.02~mol\) of acetic acid. Update the molar amounts of acetic acid and acetate ion: Moles of acetic acid: \(0.08 + 0.02 = 0.10~mol\) (amount after step 2 + additional acetic acid) Moles of acetate ion: \(0.15 - 0.02 = 0.13~mol\) Now, calculate the pH using the Henderson-Hasselbalch equation: \[pH = 4.75 + \log{\frac{0.13}{0.10}}\] Calculate the value within the logarithm function and then add it to the \(pK_a\). This will give us the final pH after adding HNO3.