Hydrochloric acid neutralizedr refers to hydrochloric acid in water being neutralized by a base, producing water and an ionic salt. In aqueous solution, hydrochloric acid behaves as a strong acid, meaning essentially complete ionization into \(\mathrm{H^+}\) and \(\mathrm{Cl^-}\).
Molecular equation and ionic description
A standard strong acid–strong base neutralization uses sodium hydroxide as the base. The balanced molecular equation is:
\[ \mathrm{HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)} \]
Strong electrolytes are represented as ions in the total ionic equation:
\[ \mathrm{H^+(aq) + Cl^-(aq) + Na^+(aq) + OH^-(aq) \rightarrow Na^+(aq) + Cl^-(aq) + H_2O(l)} \]
Net ionic equation
Spectator ions appear unchanged on both sides and cancel, leaving the net ionic form:
\[ \mathrm{H^+(aq) + OH^-(aq) \rightarrow H_2O(l)} \]
Neutralization in aqueous chemistry corresponds to consumption of \(\mathrm{H^+}\) by \(\mathrm{OH^-}\), forming liquid water. The salt (such as \(\mathrm{NaCl}\)) remains dissolved as ions for strong electrolytes.
Stoichiometric relationships in neutralization
Moles of acid and base are connected by the reaction coefficients. For \(\mathrm{HCl}\) and \(\mathrm{NaOH}\), the coefficient ratio is \(1:1\), so equivalence corresponds to equal moles:
\[ n_{\mathrm{HCl}} = n_{\mathrm{NaOH}} \]
With molarity \(C\) and volume \(V\) (in liters), moles are \(n = C \cdot V\). The equivalence condition becomes:
\[ C_{\mathrm{HCl}} \cdot V_{\mathrm{HCl}} = C_{\mathrm{NaOH}} \cdot V_{\mathrm{NaOH}} \]
pH before, at, and after equivalence
Strong acid–strong base systems have a characteristic equivalence point: at \(25^\circ\mathrm{C}\), the solution is approximately neutral because neither \(\mathrm{Na^+}\) nor \(\mathrm{Cl^-}\) hydrolyzes appreciably. The equivalence-point result is:
\[ \mathrm{pH} \approx 7.00 \quad (\text{at } 25^\circ\mathrm{C}) \]
Away from equivalence, pH follows from the excess strong acid or excess strong base. Excess-mole accounting uses the total solution volume:
| Region | Excess species | Concentration expression | pH relationship |
|---|---|---|---|
| Before equivalence | \(\mathrm{H^+}\) | \(\displaystyle [\mathrm{H^+}] = \frac{n_{\mathrm{HCl}} - n_{\mathrm{NaOH}}}{V_\text{tot}}\) | \(\displaystyle \mathrm{pH} = -\log_{10}([\mathrm{H^+}])\) |
| After equivalence | \(\mathrm{OH^-}\) | \(\displaystyle [\mathrm{OH^-}] = \frac{n_{\mathrm{NaOH}} - n_{\mathrm{HCl}}}{V_\text{tot}}\) | \(\displaystyle \mathrm{pOH} = -\log_{10}([\mathrm{OH^-}]),\ \mathrm{pH} = 14.00 - \mathrm{pOH}\) |
Titration-curve visualization
Worked numerical example
A concrete case illustrates equivalence and post-equivalence pH using hydrochloric acid neutralization by sodium hydroxide.
\(\mathrm{HCl}\): \(C_{\mathrm{HCl}} = 0.100\ \mathrm{mol\cdot L^{-1}}\), \(V_{\mathrm{HCl}} = 25.0\ \mathrm{mL}\).
\(\mathrm{NaOH}\): \(C_{\mathrm{NaOH}} = 0.125\ \mathrm{mol\cdot L^{-1}}\).
Moles of hydrochloric acid in the initial sample:
\[ n_{\mathrm{HCl}} = 0.100\ \mathrm{mol\cdot L^{-1}} \cdot 0.0250\ \mathrm{L} = 2.50 \times 10^{-3}\ \mathrm{mol} \]
Equivalence volume of sodium hydroxide follows from \(n_{\mathrm{NaOH}} = n_{\mathrm{HCl}}\):
\[ V_{\mathrm{eq}} = \frac{n_{\mathrm{HCl}}}{C_{\mathrm{NaOH}}} = \frac{2.50 \times 10^{-3}\ \mathrm{mol}}{0.125\ \mathrm{mol\cdot L^{-1}}} = 2.00 \times 10^{-2}\ \mathrm{L} = 20.0\ \mathrm{mL} \]
A post-equivalence scenario with \(30.0\ \mathrm{mL}\) of \(0.125\ \mathrm{M}\) \(\mathrm{NaOH}\) added shows excess base.
\[ n_{\mathrm{NaOH,added}} = 0.125\ \mathrm{mol\cdot L^{-1}} \cdot 0.0300\ \mathrm{L} = 3.75 \times 10^{-3}\ \mathrm{mol} \]
\[ n_{\mathrm{OH^- ,excess}} = n_{\mathrm{NaOH,added}} - n_{\mathrm{HCl}} = (3.75 - 2.50)\times 10^{-3}\ \mathrm{mol} = 1.25 \times 10^{-3}\ \mathrm{mol} \]
\[ V_{\mathrm{tot}} = 0.0250\ \mathrm{L} + 0.0300\ \mathrm{L} = 0.0550\ \mathrm{L} \]
\[ [\mathrm{OH^-}] = \frac{1.25 \times 10^{-3}\ \mathrm{mol}}{0.0550\ \mathrm{L}} = 2.27 \times 10^{-2}\ \mathrm{mol\cdot L^{-1}} \]
\[ \mathrm{pOH} = -\log_{10}(2.27 \times 10^{-2}) = 1.64 \quad\Rightarrow\quad \mathrm{pH} = 14.00 - 1.64 = 12.36 \]
Summary statement
Hydrochloric acid neutralization in water is described by a molecular equation producing salt and water and by the net ionic equation \(\mathrm{H^+ + OH^- \rightarrow H_2O}\); stoichiometric equivalence corresponds to equal moles for \(\mathrm{HCl}\) and \(\mathrm{OH^-}\), and pH depends on the excess strong acid or strong base away from equivalence.