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: Equivalence point concept

The equivalence point is reached when the moles of acid equal the moles of base. For both titrations in question 14, the stoichiometric ratio of acid - base reaction determines the volume of titrant needed to reach equivalence. Since the amounts of reactants (in terms of moles) are based on the same stoichiometry, the equivalence - point volume is the same.

Step2: pH at equivalence point

The pH at the equivalence point depends on the nature of the salt formed. In different acid - base reactions, the anions or cations of the salts can hydrolyze in water to different extents. For example, if a strong acid and a weak base react, the salt formed will have a cation that hydrolyzes to produce $H^+$ ions, making the solution acidic at the equivalence point. If a weak acid and a strong base react, the salt formed will have an anion that hydrolyzes to produce $OH^-$ ions, making the solution basic at the equivalence point.

For question 15:

Part a: Titration of $NH_3$ with $HCl$

Step1: Initial pH

$NH_3$ is a weak base. $K_b$ for $NH_3=1.8\times10^{-5}$. Let $NH_3$ be $B$. The hydrolysis reaction is $B + H_2O
ightleftharpoons BH^++OH^-$.
\[K_b=\frac{[BH^+][OH^-]}{[B]}\]
Initial $[B]=0.200\ M$. Let $x = [OH^-]=[BH^+]$. Then $1.8\times10^{-5}=\frac{x^2}{0.200 - x}\approx\frac{x^2}{0.200}$ (since $0.200\gg x$).
\[x=\sqrt{1.8\times10^{-5}\times0.200}=\sqrt{3.6\times10^{-6}}\approx1.9\times10^{-3}\ M\]
\[pOH = -\log(1.9\times10^{-3})\approx2.72\]
\[pH = 14 - pOH=11.28\]

Step2: Half - way point

At the half - way point, $[NH_3]=[NH_4^+]$. Using the Henderson - Hasselbalch equation for a base $pOH = pK_b+\log\frac{[BH^+]}{[B]}$, when $[NH_3]=[NH_4^+]$, $pOH = pK_b$.
\[pK_b=-\log(1.8\times10^{-5})\approx4.74\]
\[pH = 14 - pK_b = 9.26\]

Step3: Equivalence point

The reaction is $NH_3+HCl = NH_4Cl$. The moles of $NH_3=n = 0.010\ L\times0.200\ M = 0.002\ mol$. The volume of $HCl$ needed to reach equivalence is $V=\frac{0.002\ mol}{0.200\ M}=0.010\ L = 10.00\ mL$. The total volume at equivalence is $V_{total}=10.00\ mL + 10.00\ mL=20.00\ mL$. The concentration of $NH_4^+$ formed is $[NH_4^+]=\frac{0.002\ mol}{0.020\ L}=0.100\ M$.
$NH_4^+$ hydrolyzes: $NH_4^+
ightleftharpoons NH_3 + H^+$, $K_a=\frac{K_w}{K_b}=\frac{1.0\times10^{-14}}{1.8\times10^{-5}}\approx5.6\times10^{-10}$
Let $x = [H^+]=[NH_3]$, then $5.6\times10^{-10}=\frac{x^2}{0.100 - x}\approx\frac{x^2}{0.100}$ (since $0.100\gg x$)
\[x=\sqrt{5.6\times10^{-10}\times0.100}=\sqrt{5.6\times10^{-11}}\approx7.5\times10^{-6}\ M\]
\[pH=-\log(7.5\times10^{-6})\approx5.12\]

Part b: Titration of $H_2A$ with $NaOH$

Step1: Initial pH

$H_2A$ is a weak diprotic acid. For the first dissociation $H_2A
ightleftharpoons H^++HA^-$, $K_{a1}=10^{-4.08}$. Let $x = [H^+]=[HA^-]$, $[H_2A]=0.200\ M$.
\[K_{a1}=\frac{[H^+][HA^-]}{[H_2A]}\]
\[10^{-4.08}=\frac{x^2}{0.200 - x}\approx\frac{x^2}{0.200}\]
\[x=\sqrt{10^{-4.08}\times0.200}=\sqrt{1.05\times10^{-4}\times0.200}\approx4.6\times10^{-3}\ M\]
\[pH=-\log(4.6\times10^{-3})\approx2.34\]

Step2: First half - way point

For the first dissociation, at the half - way point $[H_2A]=[HA^-]$. Using the Henderson - Hasselbalch equation $pH = pK_{a1}=4.08$

Step3: First equivalence point

The reaction is $H_2A+NaOH = NaHA + H_2O$. The moles of $H_2A=n = 0.010\ L\times0.200\ M = 0.002\ mol$. The volume of $NaOH$ needed for the first equivalence is $V=\frac{0.002\ mol}{0.200\ M}=0.010\ L = 10.00\ mL$. The total volume is $V_{total}=10.00\ mL + 10.00\ mL = 20.00\ mL$. The concentration of $HA^-$ is $[HA^-]=\frac{0.002\ mol}{0.020\ L}=0.100\ M$.
$HA^-$ can both donate and accept a proton. The $pH$ at the first equivalence point is given by $pH=\frac{pK_{a1}+pK_{a2}}{2}=\frac{4.08 + 7.15}{2}=5.62$

Step4: Second half - way point

For the second dissociation, at the half - way point $[HA^-]=[A^{2 - }]$, $pH = pK_{a2}=7.15$

Step5: Second equivalence point

The reaction is $NaHA+NaOH = Na_2A + H_2O$. The total volume of $NaOH$ needed for the second equivalence is $20.00\ mL$. The total volume at the second equivalence is $10.00\ mL+20.00\ mL = 30.00\ mL$. The concentration of $A^{2 - }$ is $[A^{2 - }]=\frac{0.002\ mol}{0.030\ L}\approx0.067\ M$.
$A^{2 - }$ hydrolyzes: $A^{2 - }+H_2O
ightleftharpoons HA^-+OH^-$, $K_{b1}=\frac{K_w}{K_{a2}}=\frac{1.0\times10^{-14}}{10^{-7.15}}=1.41\times10^{-7}$
Let $x = [OH^-]=[HA^-]$, $[A^{2 - }]=0.067\ M$. Then $1.41\times10^{-7}=\frac{x^2}{0.067 - x}\approx\frac{x^2}{0.067}$
\[x=\sqrt{1.41\times10^{-7}\times0.067}=\sqrt{9.45\times10^{-9}}\approx9.7\times10^{-5}\ M\]
\[pOH=-\log(9.7\times10^{-5})\approx4.01\]
\[pH = 14 - pOH = 9.99\]

To sketch the curves:
- For the $NH_3 - HCl$ titration: Start at $pH = 11.28$ on the $y$ - axis ($pH$) and $x = 0$ (volume of $HCl$). At the half - way point ($V = 5.00\ mL$), $pH = 9.26$. At the equivalence point ($V = 10.00\ mL$), $pH = 5.12$.
- For the $H_2A - NaOH$ titration: Start at $pH = 2.34$. At the first half - way point ($V = 5.00\ mL$), $pH = 4.08$. At the first equivalence point ($V = 10.00\ mL$), $pH = 5.62$. At the second half - way point ($V = 15.00\ mL$), $pH = 7.15$. At the second equivalence point ($V = 20.00\ mL$), $pH = 9.99$. Label the $x$ - axis as "Volume of titrant (mL)" and the $y$ - axis as "pH".

Answer:

For question 14: The equivalence point is the same because the stoichiometric amounts of reactants determine it, but the pH is different due to the different hydrolysis behaviors of the salts formed.
For question 15: See the above - calculated pH values at key points for sketching the titration curves. The curves should be drawn with the $x$ - axis as volume of titrant and $y$ - axis as pH, and key points (initial, half - way, equivalence) labeled with their respective pH values and volumes.