Consider a buffer made by adding 132.8 g of NaC₇H₅O₂ to 300.0 mL of 1.17 M HC₇H₅O₂ (Ka = 6.3 x 10⁻⁵) What is the pH of this buffer?

Question

Consider a buffer made by adding 132.8 g of NaC₇H₅O₂ to 300.0 mL of 1.17 M HC₇H₅O₂ (Ka = 6.3 x 10⁻⁵) What is the pH of this buffer?

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Gia Bảo 3 years 2021-09-04T11:49:08+00:00 1 Answers 28 views 0

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    2021-09-04T11:50:25+00:00

    Answer:

    pH = 1.62

    Explanation:

    We can solve this problem using Henderson-Hasselbach’s equation:

    • pH = pKa + log\frac{[A^-]}{[HA]}

    Where

    • pKa = -log(Ka) = 4.20
    • [A⁻] = [NaC₇H₅O₂]
    • [HA] = [HC₇H₅O₂]

    Now we convert 132.8 g of NaC₇H₅O₂ into moles, using its molar mass:

    • 132.8 ÷ 144.11 g/mol = 0.921 moles

    Then we calculate [NaC₇H₅O₂]:

    • 0.921 moles / 300.0 mL = 0.003 M

    Now we can proceed to calculate the pH of the solution:

    • pH = 4.20 + log\frac{0.003}{1.17}
    • pH = 1.62

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