Write the net ionic equation for the following molecular equation.
3mnbr2(aq) + 2na3po4(aq) → mn3 (po4)2(s) + 6nabr(aq)

Net ionic equation:

3Mn⁺²(aq)  + 2PO₄³⁻(aq)  →  Mn₃(PO₄)₂ (s)

Explanation:

Chemical equation:

MnBr₂(aq) +  Na₃PO₄(aq)  →  Mn₃(PO₄)₂ (s) + NaBr(aq)

Balanced chemical equation:

3MnBr₂(aq) +  2Na₃PO₄(aq)  →  Mn₃(PO₄)₂ (s) + 6NaBr(aq)

Ionic equation:

3Mn⁺²(aq)  + 3Br₂²⁻(aq) +  2Na₃³⁺(aq)  + 2PO₄³⁻(aq)  →  Mn₃(PO₄)₂ (s) + 6Na⁺(aq)  + 6Br⁻(aq)

Net ionic equation:

3Mn⁺²(aq)  + 2PO₄³⁻(aq)  →  Mn₃(PO₄)₂ (s)

The Br⁻ and Na⁺ are spectator ions that's why these are not written in net ionic equation. The  Mn₃(PO₄)₂  can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.  

The question is incomplete, here is the complete question:

Calculate the pH of a solution prepared by dissolving 0.370 mol of formic acid (HCO₂H) and 0.230 mol of sodium formate (NaCO₂H) in water sufficient to yield 1.00 L of solution. The Ka of formic acid is 1.77 × 10⁻⁴

a) 2.099

b) 10.463

c) 3.546

d) 2.307

e) 3.952

The pH of the solution is 3.546

Explanation:

We are given:

Moles of formic acid = 0.370 moles

Moles of sodium formate = 0.230 moles

Volume of solution = 1 L

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

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:  

= negative logarithm of acid dissociation constant of formic acid = 3.75

pH = ?  

Putting values in above equation, we get:  

Hence, the pH of the solution is 3.546