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22 January, 22:34

What mass of solid sodium formate (of MW 68.01) must be added to 150 mL of 0.42 mol/L formic acid (HCOOH) to make a buffer solu - tion having a pH of 3.74? Ka = 0.00018 for HCOOH.

Answer in units of g.

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  1. 22 January, 22:43
    0
    We need 4.28 grams of sodium formate

    Explanation:

    Step 1: Data given

    MW of sodium formate = 68.01 g/mol

    Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L

    pH = 3.74

    Ka = 0.00018

    Step 2: Calculate [base)

    3.74 = - log (0.00018) + log [base]/[acid]

    0 = log [base]/[acid]

    0 = log [base] / 0.42

    10^0 = 1 = [base]/0.42 M

    [base] = 0.42 M

    Step 3: Calculate moles of sodium formate:

    Moles sodium formate = molarity * volume

    Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles

    Step 4: Calculate mass of sodium formate:

    Mass sodium formate = moles sodium formate * Molar mass sodium formate

    Mass sodium formate = 0.063 mol * 68.01 g/mol

    Mass sodium formate = 4.28 grams

    We need 4.28 grams of sodium formate
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