[tex]V_{NaOH}=38,1mL=0,0381L\\
C_{m}=0,265M\\\\
n=C_{m}*V=0,265\frac{mol}{L}*0,0381L\approx0,0101mol[/tex]
HR + NaOH ⇒ NaR + H₂O
1mol : 1mol
[tex]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}[/tex]
