A 1.757-g sample of a / alloy was dissolved in acid and diluted to exactly 250.0 mL in a volumetric flask. A 50.00-mL aliquot of the diluted

Question

A 1.757-g sample of a / alloy was dissolved in acid and diluted to exactly 250.0 mL in a volumetric flask. A 50.00-mL aliquot of the diluted solution was brought to a pH of 10.0 with an / buffer; the subsequent titration involved both cations and required 28.89 mL of 0.06950 M EDTA. A second 50.00-mL aliquot was brought to a pH of 10.0 with an / buffer, which also served to mask the ; 19.07 mL of the EDTA solution were needed to titrate the . Calculate the percent and in the sample.

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King 6 months 2021-08-24T23:38:52+00:00 1 Answers 12 views 0

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    2021-08-24T23:40:45+00:00

    Answer:

    78.14% Pb²⁺ and 21.86% of Cd²⁺

    Explanation:

    The first titration involves the reaction of both Pb²⁺ and Cd²⁺

    In the second titration, as the buffer is HCN/NaCN, the Cd²⁺ precipitates as Cd(CN)₂ and the only ion that reacts is Pb²⁺

    In the first titration:

    Moles EDTA = Moles Pb²⁺ and Cd²⁺:

    28.89mL = 0.02889L * (0.06950moles / L) = 2.008×10⁻³ moles in the aliquot. In the sample:

    2.008×10⁻³ moles * (250.0mL / 50.0mL) =

    0.01004 moles = Pb²⁺ + Cd²⁺ (1)

    In the second titration:

    19.07mL = 0.01907L * (0.06950mol / L) = 1.325×10⁻³ moles Pb²⁺ in the aliquot. In the sample:

    1.325×10⁻³ moles Pb²⁺ * (250.0mL / 50.0mL) =

    6.626×10⁻³ moles Pb²⁺

    That means the moles of Cd²⁺ are:

    0.01004 moles = Cd²⁺ + 6.626×10⁻³ moles Cd²⁺

    3.413×10⁻³ moles Cd²⁺

    The mass of each ion is:

    Cd²⁺ -Molar mass: 112.411g/mol-:

    3.413×10⁻³ moles Cd²⁺ * (112.411g / mol) =

    0.384g of Cd²⁺

    Pb²⁺ -Molar mass: 207.2g/mol-:

    6.626×10⁻³ moles Pb²⁺ * (207.2g / mol) =

    1.373g of Pb²⁺

    The percent mass of each ion is:

    1.373g Pb²⁺ / 1.757g = 78.14% Pb²⁺

    And:

    0.384g of Cd²⁺ / 1.757g * 100 = 21.86% of Cd²⁺

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