A titration is done using a 0.1302 M NaOH solution to determine the molar mass of a monoprotic weak acid (HA). If 50.00 mL of a 1.863 g sample of the unknown acid (HA) is titrated to the equivalence point/end point with 70.11 mL of a 0.1302 M NaOH aqueous solution, what is the molar mass of the unknown acid?
Solution:- The balanced equation for the reaction of given monoprotic acid and NaOH is...
HA(aq) + NaOH(aq) ----> NaA(aq) + H2O(l)
From the equation they react in 1:1 mol ratio. We could calculate the moles of NaOH used to react the end point from it's given volume and molarity and then using mol ratio the moles of acid could also be calculated.
for NaOH, 70.11 mL x (1L/1000 mL) x (0.1302 mol/L) = 0.009128 mol of NaOH
since the ratio is 1:1 so moles of HA used would also be 0.009128.
mass of acid is given as 1.863 g.
molar mass of acid = 1.863 g/0.009128 mol = 204.10 g/mol
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