Anonymous
Anonymous asked in Science & MathematicsChemistry · 2 weeks ago

Chemistry Help?

A solution of an unknown monoprotic weak acid, HA, has a pH of 4.50. The Ka of HA is 5.10 ✕ 10-8. What was the initial molarity of the solution of HA?

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  • 2 weeks ago
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    HA +H2O = H3O++ A-

    for each mole of H3O+ formed we get 1 mole of A-

    pH= 4.50 

    H3O+ = 10 ^-4.5 =  3.16*10^-5 = [A-]

    Ka = [H3O+] * [A-] / [HA]

    [HA] =  [H3O+] * [A-] / Ka 

    [HA] = (3.16*10^-5 ) ^2 / 5.10 ✕ 10-8

    [HA} = 0.0196 M 

    Initial concentration of HA = 0.0196 M 

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