Question

A solution containing a mixture of 0.0333 M0.0333 M potassium chromate ( K2CrO4K2CrO4 ) and 0.0532 M0.0532 M sodium oxalate ( Na2C2O4Na2C2O4 ) was titrated with a solution of barium chloride ( BaCl2BaCl2 ) for the purpose of separating CrO2−4CrO42− and C2O2−4C2O42− by precipitation with the Ba2+Ba2+ cation. The solubility product constants ( KspKsp ) for BaCrO4BaCrO4 and BaC2O4BaC2O4 are 2.10×10−102.10×10−10 and 1.30×10−61.30×10−6 , respectively.
A) Which barium salt will precipitate first? O BaC204 O BaCrO4
B) What concentration of Ba2+ must be present for BaCrO4 to begin precipitating? Number [Ba2+]=
C) What concentration of Ba is required to reduce oxalate to 10% of its original concentration? Number 2+
D) What is the ratio of oxalate to chromate (IC204CrO4) when the Ba2 concentration is 0.0050 M? Number C,O Cro,

Answer 1

Answer:

Explanation:

K₂CrO₄ + ( COONa )₂ + 2BaCl₂ = Ba CrO₄ + ( COO ) ₂ Ba + 2 KCl + 2 NaCl

.033 M             .053 M    

Ksp of   Ba CrO₄  is      2.10×10⁻¹⁰

Ksp of ( COO ) ₂ Ba  is   1.30×10⁻⁶

A ) Ksp of   Ba CrO₄ is less so it will precipitate out first .

B) Ksp = 2.10×10⁻¹⁰

Ba CrO₄ = Ba⁺²  +  CrO₄⁻²

                   C            .033

C x   .033  = 2.10×10⁻¹⁰

C =  63.63 x 10⁻¹⁰ M

Ba⁺² must be present in concentration = 63.63 x 10⁻¹⁰ M

C)

90% of precipitation of barium oxalate

concentration of oxalate to precipitate out = .9 x .0532 = .04788

( COO ) ₂ Ba  =  (COO)₂⁻²     +       Ba⁺²

                           .04788 M            C  

C x  .04788  =  1.30×10⁻⁶

C = 27.15  x 10⁻⁶ M .