Q. \[ 3\mathrm{NaOH} + \mathrm{H}_3\mathrm{PO}_4 \rightarrow \mathrm{Na}_3\mathrm{PO}_4 + 3\mathrm{H}_2\mathrm{O} \]

Q: What is the balanced reaction between \mathrm{NaOH} and \mathrm{H_3PO_4} for full neutralization?

The fully neutralized equation is \mathrm{3\,NaOH} + \mathrm{H_3PO_4} = \mathrm{Na_3PO_4} + 3\,\mathrm{H_2O} . Stoichiometry shows 3 moles of NaOH per mole of H3PO4.

Q: What is the net ionic equation for neutralization?

The common net ionic step is \mathrm{H^+} + \mathrm{OH^-} = \mathrm{H_2O} . Phosphoric acid supplies successive \mathrm{H^+} equivalents that react with \mathrm{OH^-} .

Q: How many moles of NaOH are needed to neutralize 0.100 mol of H3PO4?

For full neutralization use 3 equivalents: n_{\mathrm{NaOH}} = 3\times 0.100 = 0.300 moles. For first proton only, use 0.100 mole.

Q: How to calculate volume of NaOH solution required?

Use n = M\times V . So V = n/M . For full neutralization with 0.300 mol needed and 0.500 M NaOH, V = 0.300/0.500 = 0.600 L.

Q: Any safety or practical concerns?

\mathrm{NaOH} is strongly caustic and \mathrm{H_3PO_4} is corrosive. Use gloves, goggles, and proper ventilation. Add acid to water when diluting acids. Dispose according to local regulations.

Answer

Short explanation: Sodium hydroxide neutralizes phosphoric acid. For full neutralization, 3 moles of NaOH react with 1 mole of H3PO4 to give trisodium phosphate and water.

\[
H_3PO_4 + 3\,NaOH \rightarrow Na_3PO_4 + 3\,H_2O
\]

Detailed Explanation

We are neutralizing sodium hydroxide with phosphoric acid. The reactants are \( \mathrm{NaOH} \) and \( \mathrm{H_3PO_4} \). Phosphoric acid is a triprotic acid. That means it can donate up to three protons, one at a time. Each proton donated can be neutralized by one molecule of \( \mathrm{NaOH} \). We will write the three stepwise neutralization reactions, check that each is balanced, and then add them to obtain the overall balanced reaction.

Step 1. Neutralization of the first acidic hydrogen. One proton of \( \mathrm{H_3PO_4} \) is neutralized by one \( \mathrm{NaOH} \) to give monosodium phosphate and water. The equation is

\[ \mathrm{H_3PO_4} + \mathrm{NaOH} = \mathrm{NaH_2PO_4} + \mathrm{H_2O} \]

Check atoms on each side of this equation. Sodium count 1 equals 1. Phosphorus count 1 equals 1. Hydrogen count: left has 3 from \( \mathrm{H_3PO_4} \) plus 1 from \( \mathrm{NaOH} \) total 4, right has 2 in \( \mathrm{NaH_2PO_4} \) plus 2 in \( \mathrm{H_2O} \) total 4. Oxygen count: left has 4 in \( \mathrm{H_3PO_4} \) plus 1 in \( \mathrm{NaOH} \) total 5, right has 4 in \( \mathrm{NaH_2PO_4} \) plus 1 in \( \mathrm{H_2O} \) total 5. The equation is balanced.

Step 2. Neutralization of the second acidic hydrogen. One additional \( \mathrm{NaOH} \) reacts with \( \mathrm{NaH_2PO_4} \) to give disodium hydrogen phosphate and water. The equation is

\[ \mathrm{NaH_2PO_4} + \mathrm{NaOH} = \mathrm{Na_2HPO_4} + \mathrm{H_2O} \]

Check atoms. Sodium: left 2 equals right 2. Phosphorus: 1 equals 1. Hydrogen: left has 2 in \( \mathrm{NaH_2PO_4} \) plus 1 in \( \mathrm{NaOH} \) total 3, right has 1 in \( \mathrm{Na_2HPO_4} \) plus 2 in \( \mathrm{H_2O} \) total 3. Oxygen: left 4 plus 1 equals 5, right 4 plus 1 equals 5. Balanced.

Step 3. Neutralization of the third acidic hydrogen. One more \( \mathrm{NaOH} \) reacts with \( \mathrm{Na_2HPO_4} \) to give trisodium phosphate and water. The equation is

\[ \mathrm{Na_2HPO_4} + \mathrm{NaOH} = \mathrm{Na_3PO_4} + \mathrm{H_2O} \]

Check atoms. Sodium: left 3 equals right 3. Phosphorus: 1 equals 1. Hydrogen: left 1 plus 1 equals 2, right 2 in \( \mathrm{H_2O} \) equals 2. Oxygen: left 4 plus 1 equals 5, right 4 plus 1 equals 5. Balanced.

Now add the three stepwise equations together. When we add them, intermediate species cancel and the stoichiometric sum gives the overall neutralization of phosphoric acid by three equivalents of sodium hydroxide. The summed equation is

\[ 3\,\mathrm{NaOH} + \mathrm{H_3PO_4} = \mathrm{Na_3PO_4} + 3\,\mathrm{H_2O} \]

Final check of the overall balanced equation. Sodium: left 3 equals right 3. Phosphorus: left 1 equals right 1. Hydrogen: left 3 in \( \mathrm{H_3PO_4} \) plus 3 in \( 3\,\mathrm{NaOH} \) total 6, right 3 times 2 in \( 3\,\mathrm{H_2O} \) equals 6. Oxygen: left 4 in \( \mathrm{H_3PO_4} \) plus 3 in \( 3\,\mathrm{NaOH} \) total 7, right 4 in \( \mathrm{Na_3PO_4} \) plus 3 in \( 3\,\mathrm{H_2O} \) total 7. The overall equation is balanced.

Note about partial neutralization. If less than three moles of \( \mathrm{NaOH} \) are used per mole of \( \mathrm{H_3PO_4} \), the partially neutralized salts are the products. For one equivalent of base the product is monosodium phosphate, \( \mathrm{NaH_2PO_4} \). For two equivalents the product is disodium hydrogen phosphate, \( \mathrm{Na_2HPO_4} \). For three equivalents the fully neutralized salt \( \mathrm{Na_3PO_4} \) is formed, as given above.

See full solution

Struggling with NaOH + H3PO4? Get instant help with our AI tools.

AI homework helper

Chemistry FAQs

What is the balanced reaction between \( \mathrm{NaOH} \) and \( \mathrm{H_3PO_4} \) for full neutralization?

The fully neutralized equation is \( \mathrm{3\,NaOH} + \mathrm{H_3PO_4} = \mathrm{Na_3PO_4} + 3\,\mathrm{H_2O} \). Stoichiometry shows 3 moles of NaOH per mole of H3PO4.

What products form with partial neutralization?

Depending on equivalents of base you get: first neutralization \( \mathrm{NaOH} + \mathrm{H_3PO_4} = \mathrm{NaH_2PO_4} + \mathrm{H_2O} \). Second: \( \mathrm{2\,NaOH} + \mathrm{H_3PO_4} = \mathrm{Na_2HPO_4} + 2\,\mathrm{H_2O} \).

What is the net ionic equation for neutralization?

The common net ionic step is \( \mathrm{H^+} + \mathrm{OH^-} = \mathrm{H_2O} \). Phosphoric acid supplies successive \( \mathrm{H^+} \) equivalents that react with \( \mathrm{OH^-} \).

How many moles of NaOH are needed to neutralize 0.100 mol of H3PO4?

For full neutralization use 3 equivalents: \( n_{\mathrm{NaOH}} = 3\times 0.100 = 0.300 \) moles. For first proton only, use 0.100 mole.

How to calculate volume of NaOH solution required?

Use \( n = M\times V \). So \( V = n/M \). For full neutralization with 0.300 mol needed and 0.500 M NaOH, \( V = 0.300/0.500 = 0.600 \) L.

What are the equivalence points in titration of \( \mathrm{H_3PO_4} \) with \( \mathrm{NaOH} \)?

There are three equivalence points corresponding to neutralizing each proton. pH at each depends on the conjugate base. The first is acidic, the second near neutral to slightly basic, the third basic.

How to choose indicators for titration?

Pick indicators whose color change range overlaps the expected pH at the desired equivalence point. Phenolphthalein works for later equivalence points. Methyl orange may detect the first, acidic equivalence.

Any safety or practical concerns?

\( \mathrm{NaOH} \) is strongly caustic and \( \mathrm{H_3PO_4} \) is corrosive. Use gloves, goggles, and proper ventilation. Add acid to water when diluting acids. Dispose according to local regulations.

Explore three AI tools for chemistry.
Try them.

198,410+ active customers
Analytical, General, Biochemistry, etc.

Chemistry AI Organic Chemistry Solver Science Solver