Question

A 0.9679-g sample containing dimethylphthalate, (194.19 g/mol), and unreactive species was refluxed with 50.00 mL of 0.1215 M to hydrolyze the ester groups (this process is called saponification).

C6H4(COOCH3)2 + 2OH----->> C6H4(COO)-2 + H2O

After the reaction was complete, the excess NaOH was back titrated with 32.25mL of 0.1251M HCl. Calculate the percentage of dimethylphthalate in the sample.

Answer 1

20.44% is the percentage of dimethylphthalate in the sample.

Step-by-step explanation:

Molarity of NaOH =
`M_1=0.1215 M`

Volume of NaOH consumed in back titration =
`V_1=?`

Molarity of HCl =
`M_2=0.1251 M`

Volume of HCl =
`V_2=32.25 ml`

`M_1V_1=M_2V_2`

`V_1=(M_2V_2)/(M_1)=(0.1251 M * 50.0 mL)/(0.1215 M)`

`V_1=33.21 mL`

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)

`n=0.1215 M* 0.01679 L =0.002040 mol`

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

`C_6H_4(COOCH_3)_2 + 2OH^-\rightarrow C_6H_4(COO)^(2-) + 2H_2O`

According to reaction, 2 moles of hydroxide ion reacts with 1 mole of dimethylphthalate , then 0.002040 moles of hydroxide ion swill react with ;

`(1)/(2)* 0.002040 mol=0.001020 mol` 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;

`=(0.1979 g)/(0.9679 g)* 100=20.44\%`

20.44% is the percentage of dimethylphthalate in the sample.