Ask Question
29 March, 21:46

Consider an acetate buffer, initially at the same pH as its pKa (4.76). When a large quantity of sodium hydroxide (NaOH) is mixed with this buffer, the: A) pH remains constant. B) pH rises more than if an equal amount of NaOH is added to an acetate buffer initially at pH 6.76. C) pH rises more than if an equal amount of NaOH is added to unbuffered water at pH 4.76. D) Ratio of acetic acid to sodium acetate in the buffer falls. E) Sodium acetate formed precipitates because it is less soluble than acetic acid.

+1
Answers (1)
  1. 29 March, 22:40
    0
    D) Ratio of acetic acid to sodium acetate in the buffer falls.

    Explanation:

    A buffer is a solution that maintains the pH almost unaltered. It's formed by weak acid and its conjugate base, or a weak base and its conjugate acid. Normally we make a buffer mixing the acid or base with a salt that contains its conjugate.

    In the buffer will have an equilibrium, so by adding an acid or a base in the solution, the compounds such regulate the concentration of ions H₃O⁺ and OH⁻. The buffer has a limit: it can make the pH unaltered in pKa + / - 1. pKa is the - log Ka, where Ka is the constant of the equilibrium of the acid.

    So, adding a large quantity of sodium hydroxide, which is a base, during a certain time the buffer will work out, but the rate of equilibrium will fall and the pH will increase because there'll be no more ions to neutralize the OH⁻ of NaOH.

    If the initial pH was 6.76, the pH will increase more; if the water was unbuffered, than, the pH will increase more (there'll be no ions to neutralize OH⁻); and sodium acetate may not form a precipitate, because it's a salt formed by a strong base, so it's very soluble.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Consider an acetate buffer, initially at the same pH as its pKa (4.76). When a large quantity of sodium hydroxide (NaOH) is mixed with this ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers