What is the correct way to show sodium chloride in aqueous solution? A.NaCl (s) B.NaCl (w) C.NaCl (as) D.NaCl (aq)
2 answers:
Answer: D.NaCl (aq)
Explanation: In Sodium chloride
Sodium atom has oxidation number of +1.
Electronic configuration of sodium:
Sodium atom will loose one electron to gain noble gas configuration and form sodium cation with +1 charge.
Chlorine atom has oxidation number of -1.
Electronic configuration of sodium:
Chlorine atom will gain one electron to gain noble gas configuration and form chloride ion with -1 charge.
In sodium chloride the one electron from sodium metal gets transferred to chlorine atom to form ionic compound.
Ionic compounds dissociate to give ions when dissolved in water and thus are written as NaCl (aq).
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Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.
There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6