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A 0.125 g sample of a monoprotic acid of unknown molar mass is dissolved in water and titrated with 0.1003 M NaOH. the equivalence point is reached after adding 20.77 ml of base. What is the molar mass of the unknown acid?
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HA + NaOH + > NaA + H2O
moles of NaOH at the eq point = 0.1003 M/L x 0.0277 L = 0.00277831moles
moles of HA at the eq point = 0.00277831 moles
these have a mass of 0.125 g so g / mol = 0.125 / 0.00277831
= 44.99 g / mole .. molar mass .. possibly methanoic acid
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