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Ad libitum [116K]
3 years ago
15

A solution containing a mixture of 0.0333 M0.0333 M potassium chromate ( K2CrO4K2CrO4 ) and 0.0532 M0.0532 M sodium oxalate ( Na

2C2O4Na2C2O4 ) 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,
Chemistry
1 answer:
jasenka [17]3 years ago
3 0

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 .

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What is the conjugate acid of each of the following? What is the conjugate base of each?
Lilit [14]

Answer:

a. H₂O (conjugate acid) ; b. OH⁻ (conjugate base), H₃O⁺ (conjugate acid) ; c. H₂CO₃ (conjugate acid), CO₃⁻² (conjugate base) ; d. NH₄⁺ (conjugate strong acid) e. H₂SO₄ (conjugate acid), SO₄⁻² (conjugate base) ; f. No conjugate acid either base;  g. H₂S (conjugate acid), S⁻² (conjugate base);

h. H₄N₂ (conjugate base)

Explanation:

a.  OH⁻  +  H⁺  ⇄ H₂O

The hydroxide acts like a Bronsted Lory base, so it can catch a proton. Water will be the conjugate acid.

b. H₂O, is an amphoterus compound. It can act as an acid or a base. If it is a base, the conjugate acid is the H₃O⁺. If it is an acid, the conjugate base is the OH⁻.

c. HCO₃⁻  +  H⁺  ⇄  H₂CO₃

HCO₃⁻  +  H₂O  ⇄ CO₃⁻²  +  H₃O⁺

The bicarbonate is also amphoteric. When it catches the proton, the carbonic acid is the conjugate acid, cause it works as a base.

When the HCO₃⁻ (acid) release the proton, the conjugate base is the carbonate.

d. Ammonia is a weak base, so the conjugate strong acid is the ammonium.

NH₃ + H₂O  ⇄  NH₄⁺  +  OH⁻

e. Another amphoteric compound. The acid sulfate acts an acid and a base.

(like bicarbonate). When it is a base, the conjugate acid is the sulfuric acid, when it is an acid, the conjugate base is the sulfate.

HSO₄⁻  +  H₂O  ⇄  SO₄⁻²  +  H₃O⁺

HSO₄⁻  +  H⁺  ⇄  H₂SO₄

f. H₂O₂ does not recieve H⁺ or OH⁻, and it does not release H⁺. It is a neutral compound and it doesn't act as a base or acid.

g. HS⁻ is amphoterous.

HS⁻  +  H⁺  ⇄  H₂S

HS⁻  +  H₂O  ⇄  S⁻²  +  H₃O⁺

Same case as bicarbonate or acid sulfate.

h. H₅N₂⁺  +  H₂O  ⇄  H₄N₂  + H₃O⁺

Hidrazinium acts an acid, so, the conjugate base will be the hidrazine.

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           

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And <u>solve for n</u>:

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