Question

A 0.372 g sample of a monoprotic acid is dissolved in water and titrated with 0.210 M KOH.0.210 M KOH. What is the molar mass of the acid if 29.5 mL of the KOH solution is required to neutralize the sample?

molar mass in g/mol

Answer 1

I HA +1 KOH > KATHO mono protic Base salt Acid from equation to of HA Imole required neutralise Imole of koh koti Molarity = 0-21M volume = 29.5 ml molarity number of moles of kollOvo of koll volume (ml) > moles of koH X1000 0-21= 29.5 moles of KOH = 0.006295 moles moles of HA= moles of . o-oo 6195 KOH mass of HA moles of HA Molar mass of HA Massa HA L ooft95

- 0.372 0.006195=< . rolar mass of HA 0.372 Molar mass of 2 O-006195 HA a Go. 048 olmos