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14 March, 08:25

150 ml of 0.1 m naoh is added to 200 ml of 0.1 m formic acid, and water is added to give a final volume of 1 l. what is the ph of the final solution?

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  1. 14 March, 09:47
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    Number of moles of NaOH = V (NaOH) * M (NaOH) = 0.150 L * 0.1 moles/L = 0.015 moles

    Number of moles of formic acid, HCOOH = V (HCOOH) * M (HCOOH) = 0.200 L * 0.1 moles/L = 0.020 moles

    Here, the limiting reagent is NaOH

    The reaction is represented as:

    HCOOH + NaOH ↔HCOONa + H2O

    Moles of HCOONa formed = Moles of the limiting reagent, NaOH = 0.015 moles

    Moles of HCOOH remaining = 0.020-0.015 = 0.005 moles

    Total final volume is given as 1 L

    Therefore: [HCOOH] = 0.005 moles/1 L = 0.005 M

    [HCOONa] = 0.015/1 = 0.015 M

    pKa of HCOOH = 3.74

    As per Henderson-Hasselbalch equation

    pH = pka + log[HCOONa]/[HCOOH] = 3.74+log[0.015/0.005] = 4.22

    Therefore, pH of the final solution = 4.22
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