# Determining the Equilibrium and pH of a Mixture of Salts Let's analyze the containing: - Sodium oxalate \( \mathrm{Na_2_2O_4} \- Sodium dihydrogen phosphate, \( \mathrm{H_2PO_4}) - Calcium nitrate, \mathrm{(NO_32} \### Step : Calculate moles of each salt Given: - Mass of \( \mathrm{Na_2C_2O_4} = 2\,g \) - Mass of \mathrm{NaH_2PO_4} = 5\,g \) - Mass of \( \mathrm{Ca(NO_3)_2 = 10\,g \) - Volume of water = 100 mL = .1 L **Molar masses:** - \( \mathrm{Na_2C_2O_4} \): \( 2 \times 23 + 2 \times 12 + 4 \times 16 = 134\,g/mol \) - \( \mathrm{NaH_2PO_4} \): \( 23 + 1 \times 2 + 31 + 4 \times 16 = 119\,g/mol \) - \( \mathrm{Ca(NO_3)_2} \): \( 40 + 2 \times (14 + 3 \times 16) = 164\,g/mol \) **Calculate moles:** \[ \text{Na}_2\text{C}_2\text{O}_4: \quad n = \frac{2\,g}{134\,g/mol} \approx .0149\,mol \] \[ \text{NaH}_2\text{PO}_4: \quad n = \frac{5\,g}{119\,g/mol} \approx .042\,mol \] \[ \text{Ca(NO}_3)_2: \quad n = \frac{10\,g}{164\,g/mol} \approx .061\,mol \] --- # Step 2: Write the relevant equilibria ## 2.1 Oxalate system \[ \mathrm{C_2O_4^{2-} + H_3O^+ \leftrightarrow HC_2O_4^- + H_2O} \] with equilibrium constant \( K_{a1} \) for oxalic acid, which has: - \( pK_{a1} \approx 1.25 \) - \( pK_{a2} \approx 4.28 \) Since \( \mathrm{Na_2C_2O_4} \) is a salt of the oxalate ion, it provides \( \mathrm{C_2O_4^{2-}} \). ## 2.2 Dihydrogen phosphate system \[ \mathrm{H_2PO_4^- + H_2O \leftrightarrow H_3O^+ + HPO_4^{2-}} \] with equilibrium constant \( K_{a} \) corresponding to \( pK_{a} \approx 7.2 \). ## 2.3 Calcium nitrate \[ \mathrm{Ca(NO_3)_2} \to \mathrm{Ca^{2+}} + 2\, \mathrm{NO_3^-} \] This is a soluble salt, providing calcium ions \( \mathrm{Ca^{2+}} \). --- # Step 3: Determine initial concentrations \[ \text{Total volume} = .1\,L \] \[ [\mathrm{Na_2C_2O_4}] = \frac{.0149\,mol}{.1\,L} = .149\,M \] \[ [\mathrm{NaH_2PO_4}] = \frac{.042\,mol}{.1\,L} = .42\,M \] \[ [\mathrm{Ca(NO_3)_2}] = \frac{.061\,mol}{.1\,L} = .61\,M \] --- # Step 4: Identify dominant equilibria and potential pH - The **oxalate salt** is a weak base (since oxalate can accept \(H^+\)) but primarily acts as the source of \( \mathrm{C_2O_4^{2-}} \). - The **NaH_2PO_4** is an acidic salt, capable of donating \(H^+\), thus lowering pH. - Calcium ions can precipitate with oxalate if \( \mathrm{CaC_2O_4} \) forms, but as the salts are in solution, initial pH depends on the acid-base equilibria. --- # Step 5: Approximate pH based on dominant equilibria - \( \mathrm{NaH_2PO_4} \) is the main contributor to \(H^+\) ions because it is a weak acid. - The initial \( pH \) is mainly influenced by the dissociation of \( \mathrm{H_2PO_4^-} \): \[ \mathrm{H_2PO_4^-} \leftrightarrow \mathrm{H^+} + \mathrm{HPO_4^{2-}} \] Using the \( pK_a \approx 7.2 \): \[ \text{pH} \approx pK_{a} + \log \left( \frac{[\mathrm{HPO_4^{2-}}]}{[\mathrm{H_2PO_4^-}]} \right) \] Since initially, the solution contains mostly \( \mathrm{H_2PO_4^-} \), the pH will be around 7.2, slightly acidic due to the presence of \( \mathrm{NaH_2PO_4} \). --- # **Final Step: Approximate pH Calculation** Considering the mixture, the dominant effect is the weak acid \( \mathrm{H_2PO_4^-} \), which determines the pH: \[ \boxed{ \text{pH} \approx 7.2 } \] --- # **Summary:** - The salts dissociate into their ions. - The pH is primarily controlled by the \( \mathrm{H_2PO_4^-} \) ion, giving an approximate pH of **7.2**. - The oxalate and calcium ions are present but do not significantly alter the initial pH unless precipitation occurs. --- **Note:** For more precise pH, solving the equilibrium equations with their respective \(K_a\) values and ionic activities would be required.