How monoprotic weak acid is titrated

Reference no: EM13160083

Suppose 50.0 mL of an aqueous solution containing an unknown monoprotic weak acid is titrated with 0.200 M KOH. The titration requires 22.07 mL of the potassium hydroxide solution to reach the equivalence point. What is the concentration (in molarity) of the unknown acid solution? Here is what I think I should do but I am unsure if this is correct. 22.07 mL of .2M KOH= x moles. .2=x/.02207L, X= 4.414*10^-3 Moles of KOH. Since there are that many Moles of KOH there must be equal moles of the acid. So 4.14*10-3moles/.05L= X M of initial acid== .0828M?

Reference no: EM13160083

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How monoprotic weak acid is titrated : Since there are that many Moles of KOH there must be equal moles of the acid. So 4.14*10-3moles/.05L= X M of initial acid== .0828M?

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