Q. \[ \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H}_2\mathrm{O} \]

Q: What is the stoichiometric mole ratio between \mathrm{HCl} and \mathrm{NaOH} ?

The mole ratio is 1 to 1. One mole of \mathrm{HCl} reacts with one mole of \mathrm{NaOH} to produce one mole of \mathrm{NaCl} and one mole of \mathrm{H_2O} .

Q: How do I calculate the volume of \mathrm{NaOH} needed to neutralize given volume of \mathrm{HCl} ?

Use the dilution stoichiometry formula. M_1 V_1 = M_2 V_2 . Solve for the unknown volume or concentration using known values of M and V .

Q: What is the net ionic equation for the reaction?

The net ionic equation is \mathrm{H^+} + \mathrm{OH^-} \rightarrow \mathrm{H_2O} . Spectator ion is \mathrm{Na^+} and \mathrm{Cl^-} .

Q: What salt forms and what are its properties in solution ?

Sodium chloride, \mathrm{NaCl} , forms. It is neutral salt and strong electrolyte. In dilute aqueous solution it dissociates completely to \mathrm{Na^+} and \mathrm{Cl^-} and does not affect pH significantly at equivalence.

Q: How do I calculate the concentration of the produced \mathrm{NaCl} solution ?

Moles of \mathrm{NaCl} equal moles of limiting reactant consumed. Then concentration equals moles of \mathrm{NaCl} divided by total solution volume. Example formula: [\mathrm{NaCl}] = \dfrac{\text{moles NaCl}}{V_{\text{total}}} .

Q: What if pH is < 7 or > 7 after mixing ?

If pH < 7 acid is in excess. Calculate excess [ \mathrm{H^+} ] and pH. If pH > 7 base is in excess. Calculate excess [ \mathrm{OH^-} ] then pH = 14 - pOH. Check dilution and ionic strength for precise values.

Answer

Neutralization of hydrochloric acid with sodium hydroxide produces sodium chloride and water. Balanced molecular equation:

\[
\mathrm{HCl\,(aq)} + \mathrm{NaOH\,(aq)} = \mathrm{NaCl\,(aq)} + \mathrm{H_2O\,(l)}
\]

Complete ionic equation:

\[
\mathrm{H^+\,(aq)} + \mathrm{Cl^-\,(aq)} + \mathrm{Na^+\,(aq)} + \mathrm{OH^-\,(aq)} = \mathrm{Na^+\,(aq)} + \mathrm{Cl^-\,(aq)} + \mathrm{H_2O\,(l)}
\]

Net ionic equation:

\[
\mathrm{H^+\,(aq)} + \mathrm{OH^-\,(aq)} = \mathrm{H_2O\,(l)}
\]

Detailed Explanation

We are asked to react hydrochloric acid with sodium hydroxide. This is a simple acid–base neutralization. The balanced molecular equation is shown below.

\[ \mathrm{HCl} + \mathrm{NaOH} = \mathrm{NaCl} + \mathrm{H_{2}O} \]

Step 1. Identify reactants and products. Hydrochloric acid, \( \mathrm{HCl} \), is a strong acid. Sodium hydroxide, \( \mathrm{NaOH} \), is a strong base. The products of their neutralization are sodium chloride, \( \mathrm{NaCl} \), and water, \( \mathrm{H_{2}O} \).

Step 2. Balance the equation by counting atoms on each side. Count hydrogen, chlorine, sodium and oxygen atoms.

– Left side: hydrogen atoms: \(1\) from \( \mathrm{HCl} \) and \(1\) from \( \mathrm{NaOH} \), total \(2\). Chlorine atoms: \(1\). Sodium atoms: \(1\). Oxygen atoms: \(1\).

– Right side: hydrogen atoms: \(2\) in \( \mathrm{H_{2}O} \). Chlorine atoms: \(1\) in \( \mathrm{NaCl} \). Sodium atoms: \(1\) in \( \mathrm{NaCl} \). Oxygen atoms: \(1\) in \( \mathrm{H_{2}O} \).

All atom counts match, so the equation as written is balanced. All stoichiometric coefficients are \(1\).

Step 3. Write the ionic forms because both acid and base are strong electrolytes and fully dissociate in water. The complete ionic equation is

\[ \mathrm{H^{+}} + \mathrm{Cl^{-}} + \mathrm{Na^{+}} + \mathrm{OH^{-}} = \mathrm{Na^{+}} + \mathrm{Cl^{-}} + \mathrm{H_{2}O} \]

Step 4. Cancel spectator ions. The sodium ion, \( \mathrm{Na^{+}} \), and the chloride ion, \( \mathrm{Cl^{-}} \), appear on both sides and are spectators. Removing them yields the net ionic equation.

\[ \mathrm{H^{+}} + \mathrm{OH^{-}} = \mathrm{H_{2}O} \]

Step 5. Stoichiometry and example. The mole ratio is \(1\) mole \( \mathrm{HCl} \) reacts with \(1\) mole \( \mathrm{NaOH} \) to give \(1\) mole \( \mathrm{NaCl} \) and \(1\) mole \( \mathrm{H_{2}O} \). For example, if you mix \(0.100\) mol \( \mathrm{HCl} \) with \(0.100\) mol \( \mathrm{NaOH} \), you will produce \(0.100\) mol \( \mathrm{NaCl} \) and \(0.100\) mol \( \mathrm{H_{2}O} \).

Step 6. pH result. When equimolar amounts of a strong acid and a strong base are mixed, the resulting solution is neutral (pH about \(7\) at \(25^\circ\mathrm{C}\) ). Sodium chloride is a neutral salt and does not hydrolyze to change the pH in this ideal case.

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Chemistry FAQs

What type of reaction is \( \mathrm{HCl} + \mathrm{NaOH} \) ?

It is neutralization reaction. \[ \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O} \]

What is the stoichiometric mole ratio between \( \mathrm{HCl} \) and \( \mathrm{NaOH} \) ?

The mole ratio is 1 to 1. One mole of \( \mathrm{HCl} \) reacts with one mole of \( \mathrm{NaOH} \) to produce one mole of \( \mathrm{NaCl} \) and one mole of \( \mathrm{H_2O} \).

How do I calculate the volume of \( \mathrm{NaOH} \) needed to neutralize given volume of \( \mathrm{HCl} \) ?

Use the dilution stoichiometry formula. \( M_1 V_1 = M_2 V_2 \) . Solve for the unknown volume or concentration using known values of \( M \) and \( V \) .

How do I find the pH after mixing \( \mathrm{HCl} \) and \( \mathrm{NaOH} \) ?

Compute moles acid and base. If equal, pH is about 7. If acid excess, find \([ \mathrm{H^+} ] = \dfrac{\text{moles acid - moles base}}{\text{total volume}} \) and pH = \(-\log_{10}[ \mathrm{H^+} ]\). If base excess, find \([ \mathrm{OH^-} ]\) then pOH and pH = 14 - pOH. For strong species this is accurate.

What is the net ionic equation for the reaction?

The net ionic equation is \( \mathrm{H^+} + \mathrm{OH^-} \rightarrow \mathrm{H_2O} \) . Spectator ion is \( \mathrm{Na^+} \) and \( \mathrm{Cl^-} \) .

What salt forms and what are its properties in solution ?

Sodium chloride, \( \mathrm{NaCl} \) , forms. It is neutral salt and strong electrolyte. In dilute aqueous solution it dissociates completely to \( \mathrm{Na^+} \) and \( \mathrm{Cl^-} \) and does not affect pH significantly at equivalence.

How do I calculate the concentration of the produced \( \mathrm{NaCl} \) solution ?

Moles of \( \mathrm{NaCl} \) equal moles of limiting reactant consumed. Then concentration equals moles of \( \mathrm{NaCl} \) divided by total solution volume. Example formula: \( [\mathrm{NaCl}] = \dfrac{\text{moles NaCl}}{V_{\text{total}}} \) .

What if pH is < 7 or > 7 after mixing ?

If pH < 7 acid is in excess. Calculate excess \([ \mathrm{H^+} ]\) and pH. If pH > 7 base is in excess. Calculate excess \([ \mathrm{OH^-} ]\) then pH = 14 - pOH. Check dilution and ionic strength for precise values.

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