Question

1. using your work from #13, explain why the equivalence point is the same for both titrations but the ph is different at the equivalence point.
2. sketch the ph titration curves that would result from the titration of

a. 10.00 ml of 0.200 m nh₃ with 0.200 m hcl. (make sure to label axis and important phs should be calculated and noted (e.g. initial, equivalence, half way points))
b. 10.00 ml of 0.200 m h₂a (a generic weak acid) with 0.200 m naoh. (pka₁ = 4.08 and pka₂ = 7.15). (make sure to label axis and important phs should be calculated (e.g. initial, equivalence, half way points))

Explanation:

Step1: Analyze equivalence - point concept

The equivalence point is reached when the moles of acid equal the moles of base in a titration. For a given acid - base reaction, the stoichiometry determines the volume of titrant needed to reach the equivalence point. Since the stoichiometric ratio of the acid - base reaction is fixed for a particular reaction, the equivalence point volume is the same for the same moles of acid and base regardless of the acid or base's strength in terms of the amount of reactants needed to react completely.

Step2: Analyze pH at equivalence - point

The pH at the equivalence point depends on the nature of the salt formed. For a strong acid - strong base titration, the salt formed is neutral, so the pH = 7 at the equivalence point. For a weak acid - strong base titration, the salt formed hydrolyzes in water. The anion of the weak acid reacts with water to produce hydroxide ions, making the solution basic (pH>7). For a weak base - strong acid titration, the cation of the weak base reacts with water to produce hydronium ions, making the solution acidic (pH < 7). So, the pH at the equivalence point is different depending on the acid - base strength and the nature of the salt formed.

Step3: Titration of NH₃ with HCl (a)

Initial pH

NH₃ is a weak base. \(K_b=1.8\times10^{-5}\). For \(0.200\ M\ NH_3\), \(K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}\). Let \(x = [OH^-]\), then \(1.8\times10^{-5}=\frac{x\cdot x}{0.200 - x}\), assuming \(0.200 - x\approx0.200\), \(x=\sqrt{1.8\times10^{-5}\times0.200}\approx1.9\times10^{-3}\ M\), \(pOH = 2.72\), \(pH=11.28\)

Equivalence - point

Moles of \(NH_3=n = 0.010\ L\times0.200\ M=0.002\ mol\). Volume of \(HCl\) needed \(V=\frac{0.002\ mol}{0.200\ M}=0.010\ L = 10.00\ mL\). The total volume at equivalence is \(V_{total}=10.00\ mL + 10.00\ mL=20.00\ mL\). The salt formed is \(NH_4Cl\). The concentration of \(NH_4^+\) is \(0.100\ M\). \(K_a=\frac{K_w}{K_b}=\frac{1.0\times10^{-14}}{1.8\times10^{-5}}\approx5.6\times10^{-10}\). Let \(y = [H^+]\), \(K_a=\frac{y\cdot y}{0.100 - y}\), assuming \(0.100 - y\approx0.100\), \(y=\sqrt{5.6\times10^{-10}\times0.100}\approx7.5\times10^{-6}\ M\), \(pH = 5.12\)

Half - way point

At the half - way point, \(n_{NH_3}=n_{NH_4^+}\), \(pOH = pK_b=4.74\), \(pH = 9.26\)

Step4: Titration of \(H_2A\) with \(NaOH\) (b)

Initial pH

For a diprotic acid \(H_2A\), we consider the first dissociation. \(K_{a1}=10^{-4.08}\). Let \(z=[H^+]\), \(K_{a1}=\frac{z\cdot z}{0.200 - z}\), assuming \(0.200 - z\approx0.200\), \(z=\sqrt{10^{-4.08}\times0.200}\approx1.4\times10^{-3}\ M\), \(pH = 2.85\)

First equivalence point

Moles of \(H_2A=n = 0.010\ L\times0.200\ M = 0.002\ mol\). Volume of \(NaOH\) needed for first equivalence \(V=\frac{0.002\ mol}{0.200\ M}=0.010\ L = 10.00\ mL\). The total volume at first equivalence is \(V_{total}=10.00\ mL+10.00\ mL = 20.00\ mL\). The salt formed is \(NaHA\). \(pH=\frac{pK_{a1}+pK_{a2}}{2}=\frac{4.08 + 7.15}{2}=5.62\)

Second equivalence point

Volume of \(NaOH\) needed for second equivalence is \(20.00\ mL\). The total volume at second equivalence is \(10.00\ mL+20.00\ mL = 30.00\ mL\). The salt formed is \(Na_2A\). We consider the hydrolysis of \(A^{2 -}\), \(K_{b1}=\frac{K_w}{K_{a2}}=\frac{1.0\times10^{-14}}{10^{-7.15}}=1.4\times10^{-7}\). Let \(w=[OH^-]\), for \(0.067\ M\ A^{2 -}\) (concentration of \(A^{2 -}\) at second equivalence), \(K_{b1}=\frac{w\cdot w}{0.067 - w}\), assuming \(0.067 - w\approx0.067\), \(w=\sqrt{1.4\times10^{-7}\times0.067}\approx9.7\times10^{-5}\ M\), \(pOH = 4.01\), \(pH = 9.99\)

Half - way to first equivalence

At half - way to first equivalence, \(pH = pK_{a1}=4.08\)

Half - way between first and second equivalence

At half - way between first and second equivalence, \(pH = pK_{a2}=7.15\)

To sketch the curves:
- For \(NH_3\) - \(HCl\) titration: Start at \(pH = 11.28\), the pH decreases as \(HCl\) is added, at the equivalence point (\(V_{HCl}=10.00\ mL\)) the pH is \(5.12\), and the curve levels off as more \(HCl\) is added.
- For \(H_2A\) - \(NaOH\) titration: Start at \(pH = 2.85\), there is a break at the first equivalence point (\(V_{NaOH}=10.00\ mL\), \(pH = 5.62\)), and another break at the second equivalence point (\(V_{NaOH}=20.00\ mL\), \(pH = 9.99\)). Mark the half - way points with their respective pH values.

Answer:

The equivalence point is the same for both titrations because it is determined by the stoichiometry of the acid - base reaction (equal moles of acid and base reacting). The pH at the equivalence point is different because it depends on the nature of the salt formed (hydrolysis of the salt in water makes the solution acidic, basic or neutral). For the \(NH_3 - HCl\) titration, initial \(pH = 11.28\), equivalence \(pH = 5.12\), half - way \(pH = 9.26\). For the \(H_2A - NaOH\) titration, initial \(pH = 2.85\), first equivalence \(pH = 5.62\), second equivalence \(pH = 9.99\), half - way to first equivalence \(pH = 4.08\), half - way between first and second equivalence \(pH = 7.15\) and the curves can be sketched as described above.