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12 March, 22:26

You are provided with 300.0 mL of a buffer solution consisting of 0.200 M H3BO3 and 0.250 M NaH2BO3.

a. Calculate the pH of the buffer solution.

b. Calculate the pH of the buffer solution if 1.0 mL of 6.0 M HCl was added to it.

c. Calculate the volume of 6.0 M NaOH that must be added to raise the pH of the buffer solution to 10.00.

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  1. 12 March, 22:29
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    a. 9.34

    b. 9.06

    c. 6 mL

    Explanation:

    Part a.

    The pH of a buffer solution is given by the Henderson-Hasselbach equation:

    pH = pKa + log [A⁻] / [HA]

    where pKa is the negative log of Ka for the weak acid H₃BO₃ and can be obtained from reference tables, [A⁻] and [HA] are the concentrations of the weak conjugate base H₂BO₃⁻ and and the weak acid H₃BO₃ respectively.

    Proceeding with the calculations, we have

    Ka H₃BO₃ = 5.80 x 10⁻¹⁰

    pKa = - log (5.80 x 10⁻¹⁰) = 9.24

    [H₂BO₃⁻ ] = 0.250 M

    [H₃BO₃] = 0.200 M

    pH = 9.24 + log (0.250/0.200) = 9.34

    part b.

    When 1.0 mL of 6.0 M HCl is added to the buffer, we know that it will react with the conjugate base in the buffer doing what buffers do: keeping the pH within a small range according to the capacity of the buffer:

    H₂BO₃⁻ + H⁺ ⇒ H₃BO₃

    So lets calculate the new concentrations of acid and conjugate base after reaction and apply the Henderson equation again:

    Initial # of moles:

    H₃BO₃ = 0.300 L x 0.200 mol/L = 0.06 mol

    H₂BO₃⁻ = 0.200 L x 0.250 mol/L = 0.05 mol

    mol HCl = 0.001 L x 6.0 mol/L = 0.006 mol

    After reaction

    H₃BO₃ = 0.06 mol + 0.006 mol = 0.066 mol

    H₂BO₃⁻ = 0.05 mol - 0.006 mol = 0.044 mol

    New pH

    pH = 9.24 + log (0.044 / 0.66) = 9.06

    Note: There is no need to calculate the new concentrations since we have a quotient in the expression where the volumes cancel each other.

    Part c.

    We will be using the Henderson-Hasselbach equationm again but now to calculate ratio [H₂BO₃⁻] / [HBO₃] that will give us a pH of 10.00. Thenwe will make use of the stoichiometry of the reaction to calculate the volume of NaOH required.

    pH = pKa + log[H₂BO₃⁻]-[H₃BO₃]

    10.00 = 9.24 + log [H₂BO₃⁻]-[H₃BO₃]

    ⇒[H₂BO₃⁻] / [H₃BO₃] = antilog (0.76) = 5.75

    Initiall # moles:

    mol H₃BO₃ = 0.06 mol

    mol H₂BO₃ = 0.05 mol

    after consumption of H₃BO₃ from the reaction with NaOH:

    H₃BO₃ + NaOH ⇒ Na⁺ + H₂BO₃⁻ + H₂O

    mol H₃BO₃ = 0.06 - x

    mol H₂BO₃⁻ = 0.05 + x mol

    Therefore we have the algebraic expression:

    [H₂BO₃⁻] / [H₃BO₃] = mol H₂BO₃⁻ / mol HBO₃ = 5.75

    (again volumes cancel each other)

    0.05 + x / 0.06 - x = 5.75 ⇒ x = 0.044

    SO 0.037 mol NaOH were required, and since we know Molarity = mol / V we can calculate the volume of 6.0 M NaOH added:

    V = 0.044 mol / 6.0 mol/L = 0.0073 L

    V = 7.3 mL
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