In this item, we are simply to find the ions that may bond and are able to form a formula unit. We are also instructed to give out their name. There are numerous possible combinations of ions to form a compound. Some answers are given in the list below.
1. Na⁺ , Cl⁻ , NaCl ---> sodium chloride (this is most commonly known as table salt)
2. C⁴⁺ , O²⁻ , CO₂ ---> carbon dioxide
3. Al³+ , Cl⁻ , AlCl₃ ----> aluminum chloride
4. Ca²⁺ , Cl⁻ , CaCl₂ ---> calcium chloride
5. Li⁺ , Br⁻ , LiBr ---> lithium bromide
6. Mg³⁺ , O²⁻ , Mg₂O₃ ----> magnesium oxide
7. K⁺ , I⁻ , KI ---> potassium iodide
8. H⁺ , Cl⁻ , HCl --> hydrogen chloride
9. H⁺ , Br⁻ , HBr ----> hydrogen bromide
10. Na⁺ , Br⁻ , NaBr ---> sodium bromide
The pH a 0.25 m solution of C₆H₅NH₂ is equal to 3.13.
<h3>How do we calculate pH of weak base?</h3>
pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:
pH = pKb + log([HB⁺]/[B])
pKb = -log(1.8×10⁻⁶) = 5.7
Chemical reaction for C₆H₅NH₂ is:
C₆H₅NH₂ + H₂O → C₆H₅NH₃⁺ + OH⁻
Initial: 0.25 0 0
Change: -x x x
Equilibrium: 0.25-x x x
Base dissociation constant will be calculated as:
Kb = [C₆H₅NH₃⁺][OH⁻] / [C₆H₅NH₂]
Kb = x² / 0.25 - x
x is very small as compared to 0.25, so we neglect x from that term and by putting value of Kb, then the equation becomes:
1.8×10⁻⁶ = x² / 0.25
x² = (1.8×10⁻⁶)(0.25)
x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]
On putting all these values on the above equation of pH, we get
pH = 5.7 + log(0.67×10⁻³/0.25)
pH = 3.13
Hence pH of the solution is 3.13.
To know more about Henderson Hasselbalch equation, visit the below link:
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An electrolyte is a term used to describe a compound that can dissociate into ions as it is nothing but an ionic compound, a salt made up of a positively charged cation and negatively charged anion.
Here the correct answer is D. Since there are no hydrocarbons or any other organic compound, that do not possess partial let alone full charges, all of them can dissociate in solution to give their ions.
This allows for the solution to be able to conduct electricity.
Answer:
C. CH3COOH, Ka = 1.8 E-5
Explanation:
analyzing the pKa of the given acids:
∴ pKa = - Log Ka
A. pKa = - Log (1.0 E-3 ) = 3
B. pKa = - Log (2.9 E-4) = 3.54
C. pKa = - Log (1.8 E-5) = 4.745
D. pKa = - Log (4.0 E-6) = 5.397
E. pKa = - Log (2.3 E-9) = 8.638
We choose the (C) acid since its pKa close to the expected pH.
⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):
- pH = pKa + Log ([CH3COO-]/[CH3COOH])
∴ pH = 4.5
∴ pKa = 4.745
⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])
⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])
⇒ 0.5692 = [CH3COO-]/[CH3COOH]
∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]
⇒ 1.8 E-5 = [H3O+](0.5692)
⇒ [H3O+] = 3.1623 E-5 M
⇒ pH = - Log ( 3.1623 E-5 ) = 4.5
