Home
Class 12
CHEMISTRY
A 1.85 g sample of an arsic- containing...

A 1.85 g sample of an arsic- containing pesticide was chemically converted to `AsO_(4)^(3-)` (atomic mas of As= 74.9) and titrated with `Pb^(2+)` to form `Pb^(3) (AsO_(4))_(2)` mL. of 0.1 M `Pb^(2+)` is required to reach the equivalence point, the mass percentages fo arsenic in the pesticide sample is closest to

Promotional Banner

Similar Questions

Explore conceptually related problems

A 1.85 g sample of an arsenic- containing pesticide was chemically converted to AsO_(4)^(3-) (atomic mas of As= 74.9) and titrated with Pb^(2+) to form Pb^(3) (AsO_(4))_(2) mL. of 0.1 M Pb^(2+) is required to reach the equivalence point, the mass percentages fo arsenic in the pesticide sample is closest to

A 1.025 g sample containing a weak acid HX (mol. Mass=82) is dissolved in 60 mL. water and titrated with 0.25 M NaOH. When half of the acid was neutralised the pH was found to be 5.0 and at the equivalence point the pH is 9.0. Calculate mass precentage of HX in sample :

A 1.025 g sample containing a weak acid HX (mol. Mass=82) is dissolved in 60 mL. water and titrated with 0.25 M NaOH. When half of the acid was neutralised the pH was found to be 5.0 and at the equivalence point the pH is 9.0. Calculate mass precentage of HX in sample :

A 0.5 g sample containing MnO_(2) is treated with HCl liberating Cl_(2) is passed into a solution of KI and 30.0 " mL of " 0.1 M Na_(2)S_(2)O_(3) are required to titrate the liberated iodine. Calculate the percentage of MnO_(2) is the sample.

A 0.5 g sample containing MnO_(2) is treated with HCl liberating Cl_(2) is passed into a solution of KI and 30.0 " mL of " 0.1 M Na_(2)S_(2)O_(3) are required to titrate the liberated iodine. Calculate the percentage of MnO_(2) is the sample.

A 5.00 g sample of vinegar is titrated with 0.108 M NaOH . If the vinger requires 39.1 mL of the NaOH for complete reaction, the mass percentage of acetic acid (CH_(3)CO_(2)H) in the vinegar is

A 50 ml sample of acetic acid was titrated with 0.012 M KOH and 38.62 ml of base were required to reach equivalence point. What was the pH of the titration mixture when 19.31 ml of base had been added? [pK_(a) (acetic acid) = 4.74]

A 50 ml sample of acetic acid was titrated with 0.012 M KOH and 38.62 ml of base were required to reach equivalence point. What was the pH of the titration mixture when 19.31 ml of base had been added? [pK_(a) (acetic acid) = 4.74]