Chemical identity in water
A sodium hydrogen sulphate solution is an aqueous solution of sodium hydrogen sulfate, \(\mathrm{NaHSO_4}\). The name sodium bisulfate is commonly used for the same compound. In water, \(\mathrm{NaHSO_4}\) behaves as a strong electrolyte, so the ionic solute separates into ions:
\[ \mathrm{NaHSO_4(aq) \rightarrow Na^+(aq) + HSO_4^-(aq)} \]
The sodium ion \(\mathrm{Na^+}\) is a spectator ion in acid–base chemistry under typical conditions; the acid–base behavior comes from the hydrogen sulfate ion \(\mathrm{HSO_4^-}\).
Hydrogen sulfate \(\mathrm{HSO_4^-}\) is the conjugate base of a strong acid (\(\mathrm{H_2SO_4}\)) and still retains an acidic proton, so it can donate \(\mathrm{H^+}\) to water.
Acid–base equilibrium of \(\mathrm{HSO_4^-}\)
In water, \(\mathrm{HSO_4^-}\) acts as an acid and partially dissociates to sulfate:
\[ \mathrm{HSO_4^-(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + SO_4^{2-}(aq)} \]
The equilibrium constant for this dissociation is often denoted \(\mathrm{K_a}\) for \(\mathrm{HSO_4^-}\) (equivalently \(\mathrm{K_{a2}}\) for sulfuric acid). At \(25^\circ\text{C}\), \(\mathrm{K_{a2}}\) is on the order of \(10^{-2}\), indicating a moderately strong weak-acid behavior.
pH of a sodium hydrogen sulphate solution
A quantitative pH estimate follows from the \(\mathrm{HSO_4^-}\) dissociation equilibrium. Consider an illustrative solution prepared with formal concentration \(C\) of \(\mathrm{NaHSO_4}\) at \(25^\circ\text{C}\). Strong-electrolyte dissociation gives an initial \(\mathrm{HSO_4^-}\) concentration approximately equal to \(C\).
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| \(\mathrm{HSO_4^-}\) | \(C\) | \(-x\) | \(C-x\) |
| \(\mathrm{H_3O^+}\) | \(\approx 0\) | \(+x\) | \(x\) |
| \(\mathrm{SO_4^{2-}}\) | \(\approx 0\) | \(+x\) | \(x\) |
The equilibrium expression is
\[ K_a \;=\; \frac{[\mathrm{H_3O^+}][\mathrm{SO_4^{2-}}]}{[\mathrm{HSO_4^-}]} \;=\; \frac{x^2}{C-x} \]
Solving for \(x\) gives the hydronium concentration, \( [\mathrm{H_3O^+}] = x\), and the pH follows from
\[ \mathrm{pH} \;=\; -\log_{10}\!\big([\mathrm{H_3O^+}]\big) \;=\; -\log_{10}(x) \]
Worked numerical illustration
A concrete illustration uses \(C = 0.10\ \text{M}\) and a representative \(\mathrm{K_a} = 1.2\times 10^{-2}\) at \(25^\circ\text{C}\). The equilibrium condition becomes
\[ 1.2\times 10^{-2} \;=\; \frac{x^2}{0.10-x} \quad\Rightarrow\quad x^2 + (1.2\times 10^{-2})x - 1.2\times 10^{-3} = 0 \]
The physically meaningful root is
\[ x \;=\; \frac{-1.2\times 10^{-2} + \sqrt{(1.2\times 10^{-2})^2 + 4(1.2\times 10^{-3})}}{2} \;\approx\; 2.92\times 10^{-2}\ \text{M} \]
The corresponding pH is
\[ \mathrm{pH} \;\approx\; -\log_{10}(2.92\times 10^{-2}) \;\approx\; 1.53 \]
Speciation of sulfate in solution
The same \(x\) value provides a direct picture of sulfate speciation. For the example above, \([\mathrm{SO_4^{2-}}]=x\) and \([\mathrm{HSO_4^-}]=C-x\). The fraction as sulfate is \(x/C\), and the fraction remaining as hydrogen sulfate is \((C-x)/C\):
\[ \text{fraction as }\mathrm{SO_4^{2-}} = \frac{x}{C},\qquad \text{fraction as }\mathrm{HSO_4^-} = \frac{C-x}{C} \]
With \(C=0.10\ \text{M}\) and \(x\approx 2.92\times 10^{-2}\ \text{M}\), the distribution is about \(29\%\) \(\mathrm{SO_4^{2-}}\) and \(71\%\) \(\mathrm{HSO_4^-}\) (by concentration basis).
Visualization of species distribution
Common pitfalls
Sodium hydrogen sulfate is a salt, yet the solution is acidic because the anion \(\mathrm{HSO_4^-}\) contains a transferable proton and acts as an acid in water. Neutral salts (such as \(\mathrm{NaCl}\)) do not show this behavior because neither ion significantly reacts with water to generate \(\mathrm{H_3O^+}\) or \(\mathrm{OH^-}\).
The sodium ion \(\mathrm{Na^+}\) does not “create acidity” by itself; acidity is controlled by the \(\mathrm{HSO_4^- \rightleftharpoons SO_4^{2-}}\) equilibrium and the resulting hydronium concentration.