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Verizon [17]
3 years ago
9

2. In Experiment SOL, you investigated the solubility of oxalic acid. Sodium oxalate, Na2C2O¬4, is the sodium salt of this acid.

Categorize it as acidic, basic, or neutral in water. Does the salt dissolve, dissociate, or both in aqueous solutions? What about oxalic acid; does it dissolve, dissociate, or both in water? Explain.
Chemistry
1 answer:
AnnZ [28]3 years ago
6 0

Answer:

Sodium oxalate is a basic salt. In water it can be dissolved and dissociated.

The oxalic acid in water has two dissociations.

Explanation:

Na2C2O4 ---> 2Na+   +  C2O4-2

Sodium oxalate is the conjugate base of a weak acid. In water this salt, dissociates completely giving rise to the sodium and oxalate ions. As Na+ comes from a strong base, in water it does not produce hydrolysis while oxalate does react in water, because it takes a proton from it and it generates a basic hydrolysis releasing OH-.

C2O4-2  + H2O ⇄  HC2O4-  +  OH-

In water the salt is basic.  The pH of an aqueous solution of this salt is basic, since OH- is generated.

The HC2O4- has a second hydrolisis, it takes another proton from water to form oxalic acid.

HC2O4-  +  H2O ⇄  H2C2O4  +  OH-

The oxalic acid acts as a weak acid, it can release 2 protons to water, to make oxalate (its conjugate base).

H2C2O4  + H2O ⇄ H3O+  + HC2O4-

HC2O4-  +  H2O ⇄  H3O+  C2O4-2

The  HC2O4-  acts as an ampholyte since it accepts and delivers protons simultaneously.

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Will the ph of a solution of nh4cn be >7, <7, or =7?
never [62]
PH of a solution will be <span>higher than 7
</span>
Ammonium cyanide is a salt formed by hydrogen cyanide and ammonia. Ammonia is a weak base and hydrogen cyanide is a weak acid. 
NH₄CN + H₂O ⇒ NH₃ + HCN 

NH₄⁺ + H₂O -----> H₃O⁺ + NH₃

CN⁻ + H₂O -----> HCN + OH⁻ 

Although both compounds are weak electrolytes, NH₃ is somewhat stronger base than HCN is a strong acid, so the solution reacts alkaline. We can prove this using Ka and Kb values:

Ka(HCN) = 4.9 x × 10⁻¹⁰

Kb(NH₃) = 1.8 × 10⁻⁵<span>
Kw= </span>1.0 × 10⁻¹⁴

Let's first calculate Ka for NH₄⁺: 
Ka(NH₄⁺) x Kb(NH₃<span>) = pKw

</span>Ka(NH₄⁺) = Kw/Kb(NH₃) = 5.6 x 10⁻¹⁰

Then, Kb for CN⁻:

Kb(CN⁻) x Ka(HCN) = pKw

Kb(CN⁻) = Kw/Ka(HCN) = 2 x 10⁻⁵

 
From this, we can see that the acid constant NH4⁺ is much lower than the base constant of CN⁻, which will say that the solution of NH₄CN will react slightly alkaline because of the higher presence of hydroxyl ions in solution.


7 0
3 years ago
If you need to measure the volume of liquid in a bottle of eye drops, what unit would be the most practical?
aniked [119]
The volume of eye drop fluid in a bottle is specified in milliliter, because in medical 1 drop=12ml
7 0
3 years ago
3. Gasoline, coal, batteries, and logs all have chemical energy. When they burn that chemical energy will be transformed into en
Arturiano [62]

Answer:

Thermal energy

Explanation:

When gasoline, coal, batteries and logs are all burn they transform chemical energy to thermal energy.

The chemical energy is the energy held between chemical chains and bonds within an atom.

  • When they combust, they release thermal energy
  • Chemical energy is a potential energy.
  • The thermal energy is a kinetic energy
  • It increase the average motion of the particles in the medium
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3 0
3 years ago
A pharmacist has calculated that a patient requires 30 mmol of phosphate and 80 meq of potassium to be added to the pn. how many
aalyn [17]

Since potassium and phosphate is what we are to find for and they are both found in the potassium phosphate solution, therefore we solve for this one first on the basis of the phosphate.

The formula for finding the volume given the concentration and number of moles is:

Volume = number of moles / concentration in Molarity

Volume potassium phosphate required = 30 mmol phosphate / (3 mmol / mL)

<u>Volume potassium phosphate required = 10 mL</u>

This would also contain potassium in amounts of:

Amount of potassium in potassium phosphate = 10 mL (4.4 meg / mL)

Amount of potassium in potassium phosphate = 44 meg

 

Therefore the potassium chloride required is:

Volume of potassium chloride = (80 meg – 44 meg) / (2 meg / mL)

<span><u>Volume of potassium chloride = 72 mL</u></span>

8 0
3 years ago
I’LL MAKE YOU BRAINLIEST+ FREE POINTS
Pavel [41]

Answer:

The solution's new volume is 1.68 L

Explanation:

Dilution is the procedure to prepare a less concentrated solution from a more concentrated one, and simply consists of adding more solvent. So, in a dilution the amount of solute does not vary, but the volume of the solvent varies.

In summary, a dilution is a lower concentration solution than the original.

The way to do the calculations in a dilution is through the expression:

Ci*Vi=Cf*Vf

where C and V are concentration and volume, respectively; and the i and f subscripts indicate initial and final respectively.

In this case, being:

  • Ci= 7 M
  • Vi= 0.60 L
  • Cf= 2.5 M
  • Vf=?

Replacing:

7 M*0.60 L= 2.5 M* Vf

Solving:

Vf=\frac{7 M*0.60 L}{2.5 M}

Vf= 1.68 L

<u><em>The solution's new volume is 1.68 L</em></u>

4 0
3 years ago
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