it takes 23.90 mL of 0.139 N NaOH to neutralize .268 grams of an unknown acid. Calculate the equivalent mass of the acid.
You can't calculate the molar mass because the acid may contain more than 1 replacable H+. That is why you call it the equivalent mass.
23.90 mL = 0.0239 L
0.0239 L x 0.139 M = 0.003322 moles NaOH
0.003322 moles of NaOH = 0.003322 moles of H+ ions.
0.268 g / 0.003322 moles = 80.67 equivalent mass
no of equivalents of NaOH used =0.139*0.0239= 0.003322
therefore, no of equivalents of acid = .003322
no of equivalents * equivalent weight = mass of acid
equivalent mass = 0.268/.00322 = 83.229g
Comments
You can't calculate the molar mass because the acid may contain more than 1 replacable H+. That is why you call it the equivalent mass.
23.90 mL = 0.0239 L
0.0239 L x 0.139 M = 0.003322 moles NaOH
0.003322 moles of NaOH = 0.003322 moles of H+ ions.
0.268 g / 0.003322 moles = 80.67 equivalent mass
no of equivalents of NaOH used =0.139*0.0239= 0.003322
therefore, no of equivalents of acid = .003322
no of equivalents * equivalent weight = mass of acid
equivalent mass = 0.268/.00322 = 83.229g