Math Problem Statement

Calculate the pH of a 1.00 L buffer solution containing 0.153 mol of acetic acid (HC2H3O2) and 0.194 mol of sodium acetate (NaC2H3O2). The Ka of acetic acid is 1.8 x 10^-5. Calculate the pH of the buffer solution after 0.0388 mol of HCl is added to the buffer. Assume the volume of the solution does not change.

Solution

To calculate the pH of the buffer solution after the addition of 0.03880.0388 mol of HCl, we can use the Henderson-Hasselbalch equation and consider the effect of the added HCl on the buffer system.

Initial Information:

  • Initial moles of acetic acid (HC2H3O2\text{HC}_2\text{H}_3\text{O}_2): 0.1530.153 mol
  • Initial moles of acetate (NaC2H3O2\text{NaC}_2\text{H}_3\text{O}_2): 0.1940.194 mol
  • KaK_a of acetic acid = 1.8×1051.8 \times 10^{-5}
  • Initial pH calculated: 4.85

Step-by-Step Solution:

  1. Determine the effect of adding HCl:

    • HCl\text{HCl} is a strong acid and will dissociate completely, providing 0.03880.0388 mol of H+\text{H}^+.
    • The H+\text{H}^+ ions from HCl\text{HCl} will react with the acetate ions (C2H3O2\text{C}_2\text{H}_3\text{O}_2^-) in the buffer, forming additional acetic acid (HC2H3O2\text{HC}_2\text{H}_3\text{O}_2).
  2. Calculate the new amounts of acetic acid and acetate:

    • After adding 0.03880.0388 mol of HCl\text{HCl}:

      • Acetate ions (C2H3O2\text{C}_2\text{H}_3\text{O}_2^-) will decrease by 0.03880.0388 mol.
      • Acetic acid (HC2H3O2\text{HC}_2\text{H}_3\text{O}_2) will increase by 0.03880.0388 mol.
    • New moles of acetic acid = 0.153+0.0388=0.19180.153 + 0.0388 = 0.1918 mol

    • New moles of acetate = 0.1940.0388=0.15520.194 - 0.0388 = 0.1552 mol

  3. Use the Henderson-Hasselbalch equation to find the new pH: [ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) ] where:

    • pKa=log(1.8×105)4.74\text{p}K_a = -\log(1.8 \times 10^{-5}) \approx 4.74
    • [A]=0.1552[\text{A}^-] = 0.1552 mol (acetate concentration)
    • [HA]=0.1918[\text{HA}] = 0.1918 mol (acetic acid concentration)
  4. Plug in the values: pH=4.74+log(0.15520.1918)\text{pH} = 4.74 + \log \left( \frac{0.1552}{0.1918} \right) pH=4.74+log(0.809)\text{pH} = 4.74 + \log(0.809) pH=4.740.092\text{pH} = 4.74 - 0.092 pH4.65\text{pH} \approx 4.65

Final Answer:

The pH of the buffer solution after the addition of 0.03880.0388 mol of HCl is approximately 4.65.

Would you like further details or explanations on this solution?

Here are some related questions you might find helpful:

  1. How does the Henderson-Hasselbalch equation apply to buffer solutions?
  2. What is the effect of adding a strong acid to a buffer solution?
  3. How would the buffer capacity affect the pH change upon acid addition?
  4. What are some real-life applications of buffer solutions in chemistry?
  5. How does the pKa value influence the pH of a buffer?

Tip: Remember, buffers work best when the pH is close to the pKa of the acid, within about one pH unit above or below.

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Math Problem Analysis

Mathematical Concepts

Buffer solution
Acid-base equilibrium
Henderson-Hasselbalch equation

Formulas

Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA])

Theorems

Henderson-Hasselbalch equation for buffer solutions

Suitable Grade Level

Undergraduate (Chemistry)