? asked in Science & MathematicsChemistry · 2 months ago

pH problem?

In an experiment,20ml of 0.06M of methanoic acid (HCOOH) solution was pipetted into a beaker. The pH of the solution was measured and found to be pH 2.50

a) Calculate the ‘actual’ concentrations of

i) Undissociated methanoic acid[ans 0.05684 M]ii) Dissociated methanoate ions (HCOO-(aq))[ans 0.00316 M]b) Calculate the ionisation constant Kafor methanoic acid.[ans 1.76 x 10-4]

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  • Bobby
    Lv 7
    2 months ago

    HCOOH = H+ + HCOO-

    for each mole of H+ formed we get a mole of HCOO-

    pH= 2.50; H+ =  10^-2.5 

    = 0.00316 M = [HCOO-] 

    Undissociated methanoic acid = 0.06 -  0.00316 = 0.0568 M

    Ka =[H+]*[HCOO-] /[HCOOH]

    =  0.00316^2 /  0.0568

    = 1.76 * 10^-4

    there are your answers 

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