Ask Question
28 November, 01:37

how many mililitres of 0.800M KOH should be added to 5.02g of 1,2-pentanedoic acid (C5H8O4 FM 132.11) to give a pH 4.40 when diluted to 250ml

+5
Answers (1)
  1. 28 November, 01:49
    0
    25.4mL of KOH 0.800M must be added to obtain the desire pH

    Explanation:

    pKa of 1,2-pentanedioic acid is 4.34.

    It is possible to find pH of a buffer by using H-H equation, thus:

    pH = pka + log [A⁻] / [HA]

    Where [HA] is concentration of acid and [A⁻] is concentration of conjugate acid.

    Replacing:

    4.40 = 4.34 + log [A⁻] / [HA]

    1.148 = [A⁻] / [HA] (1)

    5.02g of 1,2-pentanedioic acid (Molar mass: 132.11g/mol) are:

    5.02g ₓ (1mol / 132.11g) = 0.0380 moles of acid. That means:

    [A⁻] + [HA] = 0.0380 (2)

    Replacing (1) in (2):

    1.148 = 0.0380 - [HA] / [HA]

    1.148[HA] = 0.0380 - [HA]

    2.148[HA] = 0.0380

    [HA] = 0.0177 moles

    Thus:

    [A⁻] = 0.0380 - 0.0177 = 0.0203 moles [A⁻]

    The moles of A⁻ comes from the reaction of the weak acid with KOH, that is:

    HA + KOH → A⁻ + H₂O + K⁺

    Thus, you need to add 0.0203 moles of KOH to produce 0.0203 moles of A⁻. As KOH solution is 0.800M:

    0.0203 moles KOH ₓ (1L / 0.800mol) = 0.0254L of KOH 0.800M =

    25.4mL of KOH 0.800M must be added to obtain the desire pH
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “how many mililitres of 0.800M KOH should be added to 5.02g of 1,2-pentanedoic acid (C5H8O4 FM 132.11) to give a pH 4.40 when diluted to ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers