Respuesta :
Answer:
20.44% is the percentage of dimethylphthalate in the sample.
Explanation:
Molarity of NaOH = [tex]M_1=0.1215 M[/tex]
Volume of NaOH consumed in back titration = [tex]V_1=?[/tex]
Molarity of HCl =[tex]M_2=0.1251 M[/tex]
Volume of HCl = [tex]V_2=32.25 ml[/tex]
[tex]M_1V_1=M_2V_2[/tex]
[tex]V_1=\frac{M_2V_2}{M_1}=\frac{0.1251 M \times 50.0 mL}{0.1215 M}[/tex]
[tex]V_1=33.21 mL[/tex]
Volume of NaOH used in hydrolysis of ester = 50.00 mL - 33.21 mL = 16.79 mL[/tex]
Moles of NaOH in 16.79 ml of 0.1215 M solution : n
Volume of solution = 16.79 mL = 0.01679 L ( 1mL=0.001 L)
[tex]n=0.1215 M\times 0.01679 L =0.002040 mol[/tex]
1 mole of NaOH has 1 mole of hydroxide ions than 0.002040 moles of NaOH will have :
1 × 0.002040 mol = 0.002040 mol of hydroxide ions
Moles of hydroxide ions = 0.002040 mol
[tex]C_6H_4(COOCH_3)_2 + 2OH^-\rightarrow C_6H_4(COO)^{2-} + 2H_2O[/tex]
According to reaction, 2 moles of hydroxide ion reacts with 1 mole of dimethylphthalate , then 0.002040 moles of hydroxide ion swill react with ;
[tex]\frac{1}{2}\times 0.002040 mol=0.001020 mol[/tex] of dimethylphthalate
Mass of 0.001020 moles of dimethylphthalate :
194 g/mol × 0.001020 mol = 0.1979 g
Mass of sample = 0.9679 g
Mass of dimethylphthalate = 0.1979 g
Percentage of dimethylphthalate in sample;
[tex]=\frac{0.1979 g}{0.9679 g}\times 100=20.44\%[/tex]
20.44% is the percentage of dimethylphthalate in the sample.
