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The formula weight of an acid is 82. In a titration, 100cm3 of the solution of this acid containing 39g of this acid per litre were completely neutralized by 95 cm3 of aqueous naoh containing 40g of naoh per litre. What is the basicity of this acid?

Bob helps me

  • chemistry -

    mols NaOH in solution = grams/formula weight = 40/40 = 1 mol.
    M NaOH = 1 mol/L = 1M

    mols acid = grams/formula weight = 39/82 = about 0.48 but you do it more accurately.
    M base = mols/L = 0.48M

    mols acid x #H = mols base
    0.48M x 0.100L x #H = 1M x 0.095L
    #H = (1M x 0.095)/(0.48 x 0.095)
    #H = ?

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