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Vilka [71]
3 years ago
7

Given the ion C2O4-2, what species would you expect to form with each of the following ions?

Chemistry
1 answer:
Ksivusya [100]3 years ago
6 0

Answer:

A. K₂C₂O₄          Potassium oxalate

B. CuC₂O₄          Copper oxalate

C. Bi₂(C₂O₄)₃         Bismuth (III) oxalate

D. Pb(C₂O₄)₂         Lead (IV) oxalate

E. (NH₄)₂C₂O₄       Ammonium oxalate

F. HC₂O₄⁻             Acid oxalate

Explanation:

C₂O₄⁻²  → oxalate anion

This is the conjugate base from the H₂C₂O₄ which is the oxalic acid. A weak dyprotic acid that can release 2 protons.

A. 2K⁺  +  C₂O₄⁻²  → K₂C₂O₄          Potassium oxalate

It can be formed by the neutralization of the acid with the base

H₂C₂O₄  + 2KOH  → K₂C₂O₄  +  2H₂O

B. Cu²⁺ +  C₂O₄⁻²   ⇄  CuC₂O₄  ↓

This is a precipitate.

C.  2Bi³⁺  +  3C₂O₄⁻²   ⇄  Bi₂(C₂O₄)₃  ↓

This is a precipitate.

D.  Pb⁴⁺ +  2C₂O₄⁻²   ⇄  Pb(C₂O₄)₂  ↓

This is a precipitate.

E. 2NH₄⁺  +  C₂O₄⁻²   ⇄  (NH₄)₂C₂O₄  ↓

This is a precipitate.

F. This is the conjugate strong base, for the weak acid

H⁺  +  C₂O₄⁻²   ⇄  HC₂O₄⁻

HC₂O₄⁻  + H₂O  ⇄  C₂O₄⁻²  +  H₃O⁺    Ka

HC₂O₄⁻  + H₂O  ⇄  H₂C₂O₄  +  OH⁻    Kb

HC₂O₄⁻  is an amphoteric compound

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Answer:

Explanation:

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0.065 mol I2 x 254g I₂/ 1 mol I₂  = 16.5 g I2

8 0
3 years ago
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A chemist must prepare 300.0mL of nitric acid solution with a pH of 0.70 at 25°C. He will do this in three steps: Fill a 300.0mL
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<u>Answer:</u> The volume of concentrated solution required is 9.95 mL

<u>Explanation:</u>

To calculate the pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

pH = 0.70

Putting values in above equation, we get:

0.70=-\log[H^+]

[H^+]=10^{-0.70}=0.199M

1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.

Molarity of nitric acid = 0.199 M

To calculate the volume of the concentrated solution, we use the equation:

M_1V_1=M_2V_2

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M_1\text{ and }V_1 are the molarity and volume of the concentrated nitric acid solution

M_2\text{ and }V_2 are the molarity and volume of diluted nitric acid solution

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M_1=7.0M\\V_1=?mL\\M_2=0.199M\\V_2=350mL

Putting values in above equation, we get:

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3 years ago
1. Which of the following best describes the relationship
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3 years ago
determine the ph of a buffer that is 0.55 M HNO2 and 0.75 M KNO2. tha value of Ka for HNO2 is 6.8*10^-4
Mariana [72]

Answer:

pH = 3.3

Explanation:

Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.

In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].

Solution using the I.C.E. table:

              HNO₂ ⇄    H⁺   +   KNO₂⁻

C(i)        0.55M       0M      0.75M

ΔC            -x            +x          +x

C(eq)  0.55M - x       x     0.75M + x    b/c [HNO₂] / Ka > 100, the x can be                                    

                                                             dropped giving ...

           ≅0.55M        x       ≅0.75M        

Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]

=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M

pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3

Solution using the Henderson-Hasselbalch Equation:

pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]

= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]

= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3

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