Write the Net Ionic Equation for: Hf (Aq)+naoh (Aq)ã¢â€ €™naf (Aq)+h2o (L)

1. Which net ionic equation describes the reaction between K3PO4(aq) when it is mixed with Al(NO3)3(aq)? O AB+(aq) +3NO3- (aq) – 3AINO3(s)
O K+(aq) + NO3(aq) - KNO3 (aq)
O 3K+(aq) + PO43-(aq) – 3KPO4(s)
O A13+(aq) + PO43-(aq) - AlPO4(s)​

Answers

Al (s) + 3Fe3+ (aq) ---> Al3+ (aq) + 3Fe2+(aq)

Explanation:

atoms become positive (or less negative) ions when they lose electrons. They become negative (or less positive) ions when they gain electrons, since each electron (e-) has a charge of -1.

So, since Al changes to Al3+ which is positive, it loses electrons. And when Fe3+ changes to Fe2+, it becomes less positive, so it gained electrons.

Now, write the half (balanced) equations:

Al ---> Al3+ + 3e-

Fe3+ + e- > Fe2+

To combine both equations, we have to eliminate electrons. To do so, make sure both half equations have the same no. of electrons. We can multiply the equation of Fe by 3 so that it has 3 electrons.

3Fe3+ + 3e- > 3Fe2+

Finally, we can combine both equations (remember to eliminate the electrons).

Al (s) + 3Fe3+ (aq) ---> Al3+ (aq) + 3Fe2+(aq)

Check to see if both sides of the equation have equal charges. Ions always have an aqueous state.

Al(s) + Fe³⁺ > Al³⁺(aq)  + Fe²⁺(aq)

Explanation:

The reaction between aluminium metal with an aqueous solution of Fe³⁺ ions to produce Al³⁺ ions and Fe²⁺ ions is a redox reaction in which the aluminium metal is oxidized while the Fe³⁺ ions are reduced.

Oxidation occurs if the oxidation number of an element or ionic species increases in the positive direction whereas reduction occurs when the oxidation number of the reacting species decreases.

The oxidation number of aluminium increases from 0 to +3 whereas  the oxidation number of iron reduces from +3 to =2 in the given reaction.

In a net ionic equation, only the species participating in the reaction are shown.

The equation is given below;

Al(s) + Fe³⁺ > Al³⁺(aq)  + Fe²⁺(aq)

OH- (aq) + H^+ (aq) H2O(I)

OH- + H+ H2O

Explanation:

A P E X

OH–( aq) + H+( aq) → H2O( l)

Explanation:

Step 1: Data given

Hydrochloric acid = HCl

⇒ dissociates in H+ and Cl-

Sodium hydroxide = NaOH

⇒ dissociates in Na+ and OH-

Step 2: The equation

NaOH(aq) + HCl(aq)  → Na+ + OH- + H+ +Cl-

Na+(aq)  + OH-(aq)  + H+(aq)  + Cl-(aq)  = Na+ (aq) + Cl-(aq)  + H20(l)

After canceling the spectators  ion (= ions that do not take place in a chemical equation, Na+ and Cl-)

OH–( aq) + H+( aq) → H2O( l)

Pb⁺²(aq) + CO₃⁻²(aq) →  PbCO₃ (s)

Explanation:

Pb⁺²(aq) + CO₃⁻²(aq) →  PbCO₃ (s)

In net ionic equation we cancel the ions that have equal moles on both sides so  Na⁺¹ and NO₃⁻¹ have equal moles on both sides so we canceled them.

Explanation:

Net ionic equation:

In net ionic equation we only write the ions that are involved in reaction. If the system have same moles of ions in initial and final stages we cancel them as they have the same amount and are present in ionic form in the reaction medium. To formulate an ionic equation we just cancel the ions which have the same moles in initial and final stages.

Chemical equation:

Pb(NO₃)₂ (aq) + Na₂CO₃(aq) → PbCO₃ (s) + NaNO₃ (aq)

Balanced chemical equation:

In a balanced chemical equation we write the reactants and products in molecular form with number of moles.

Pb(NO₃)₂ (aq) + Na₂CO₃(aq) → PbCO₃ (s) + 2NaNO₃ (aq)

Ionic equation:

In ionic equation we write the equation in ionic form. It involves all the ions which will produce when we add any ionic compound in reaction medium.

Pb⁺² +2NO₃⁻¹ + CO₃⁻² + 2Na⁺¹  → PbCO3 (s) + 2NO₃⁻¹ (aq) + 2Na⁺¹ (aq)

Net ionic equation

In net ionic equation we cancel the ions that have equal moles on both sides. As we can see in the above ionic equation that Na⁺¹ and NO₃⁻¹ have equal moles on both sides so we canceled them.

Pb⁺²(aq) + CO₃⁻²(aq)  → PbCO₃ (s)

the net equation that describe the reaction  between NaOH(aq) and HCL(aq)  is as below

write the  chemical equation

NaOH(aq) +HCl  (aq)  = NaCl(aq)  + H2O (l)

ionic equation

Na^+(aq)  + OH^-(aq)  + H^+(aq)  + Cl^-(aq)  = Na^+ (aq) + Cl^-9aq)  +H20(l)

cancel the spectators  ion (ion that do not take place in a chemical equation )

that  is Na^+  and  Cl^- ions

the  net ionic equation is therefore

= H^+(aq)  + OH^-(aq)  =  H20 (l)

HCl + OH- > H2O + Cl-

Neutralization reaction

K₃PO₄(aq) reacts with Al(NO₃)₃(aq) and form AlPO₄(s) and KNO₃(aq) as the products.

Before writing net ionic reaction, following steps should be followed.

Step 1 : Write the balanced chemical equation for the reaction and indicate the state of compound as aq,s,g or l.
      K₃PO₄(aq) + Al(NO₃)₃(aq) → AlPO₄(s) + 3KNO₃(aq)

Step 2 : Identify the ionic species which dissolve in water to form ions.
K₃PO₄(aq) → 3K⁺(aq) + PO₄⁻(aq)
Al(NO₃)₃(aq) → Al³⁺(aq) + 3NO₃⁻(aq)
KNO₃(aq) → K⁺(aq) + NO₃⁻(aq)

Step 3 : Write the equation again by using ions.
3K⁺(aq) + PO₄⁻(aq) + Al³⁺(aq) + 3NO₃⁻(aq) → AlPO₄(s) +3K⁺(aq) + 3NO₃⁻(aq)

Step 4 : Identify the similar species in both side of the equation and cut off them.
3K⁺(aq) and 3NO₃⁻(aq) present in both sides. Hence, those ions can be cut off.

Step 5 : Get the final ionic equation.
The net ionic equation for the given reaction is
      Al³⁺(aq) + PO₄⁻(aq) → AlPO₄(s)

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Write the Net Ionic Equation for: Hf (Aq)+naoh (Aq)ã¢â€ Â€™naf (Aq)+h2o (L)

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