Question

A 2.10g of unknown monoprotic acid is titrated with 38.10 mL of .265M NaOH. Calculate the molar mass of the ac? I already calculated that there are .0101 moles of NaOH so that I have .0101 mol of the unknown acid.

Answer 1

$V_{NaOH}=38,1mL=0,0381L\\ C_{m}=0,265M\\\\ n=C_{m}*V=0,265\frac{mol}{L}*0,0381L\approx0,0101mol$

HR    +      NaOH ⇒ NaR + H₂O
1mol   :     1mol

$0,0101mol \ \ \ \ \Rightarrow \ \ \ \ 2,1g\\ 1mol \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ m\\\\ m=\frac{1mol*2,1g}{0,0101mol}\approx 207,92g\\\\ M_{HR}=207,92\frac{g}{mol}$