33. Based on the information characterized by the curves in the graphs, the wrong conclusion is drawn.
Figure 1
Figure 2
Figure 3
Figure 4
- A. A. Figure 1 shows the pH change curve of adding $0.1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { CH } _ { 3 } \mathrm { COOH }$ drop by drop to a NaOH solution of volume $10 \mathrm {~mL} 0.1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ at room temperature, then point c has: $\mathrm { c } \left( \mathrm { CH } _ { 3 } \mathrm { COOH } \right) + 2 \mathrm { c } \left( \mathrm { H } ^ { + } \right) = 2 \mathrm { c } \left( \mathrm { OH } ^ { - } \right) + \mathrm { c } \left( \mathrm { CH } _ { 3 } \mathrm { COO } ^ { - } \right)$
- B. B. In Fig. 2, after mixing $\mathrm { pH } = \mathrm { a }$'s $\mathrm { H } _ { 2 } \mathrm { SO } _ { 4 }$ solution with $\mathrm { pH } = \mathrm { b }$'s NaOH solution in equal volume at the temperature corresponding to the point b, the solution is neutral, then $\mathrm { a } + \mathrm { b } = 12$
- C. C. From the curve in Fig. 3, $\mathrm { K } _ { \mathrm { sp } } ( \mathrm { AgCl } ) > \mathrm { K } _ { \mathrm { sp } } ( \mathrm { AgBr } ) > \mathrm { K } _ { \mathrm { sp } } ( \mathrm { AgI } )$ can be determined, so when $0.0100 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { AgNO } _ { 3 }$ standard solution is used to titrate a mixed solution with concentrations of $0.1000 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { Cl } ^ { - } , \mathrm { Br } ^ { - }$ and $\mathrm { I } ^ { - }$, the first to precipitate is $\mathrm { I } ^ { - }$. FORMULA_4]]
- D. D. Figure 4 shows the pH change curves of solutions of hydrochloric acid and acetic acid of the same pH when diluted with water, where I denotes hydrochloric acid and II denotes acetic acid, and the conductivity of the solutions: $b > a > c$
Answer: D
Solution: A. Figure 1 shows the pH change curve of adding 0.1 $\mathrm { mol } \cdot \mathrm { L } ^ { - 1 } \mathrm { CH } _ { 3 } \mathrm { COOH }$ drop by drop to a NaOH solution with a volume of $10 \mathrm {~mL} 0.1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ at room temperature, and the amount of acid added at point c is two times that of the base, and there is charge conservation $c \left( \mathrm { Na } ^ { + } \right) + c \left( \mathrm { H } ^ { + } \right) = c \left( \mathrm { OH } ^ { - } \right) + c \left( \mathrm { CH } _ { 3 } \mathrm { COO } ^ { - } \right)$ and material conservation in the solution. $2 c \left( \mathrm { Na } ^ { + } \right) = c \left( \mathrm { CH } _ { 3 } \mathrm { COO } ^ { - } \right) + c \left( \mathrm { CH } _ { 3 } \mathrm { COOH } \right)$, then at point c there is: $c \left( \mathrm { CH } _ { 3 } \mathrm { COOH } \right) + 2 c \left( \mathrm { H } ^ { + } \right) = 2 c \left( \mathrm { OH } ^ { - } \right) + c \left( \mathrm { CH } _ { 3 } \mathrm { COO } ^ { - } \right)$, so A is correct;
B. Figure 2 shows the ion product constant of water at the temperature $K _ { w } = c \left( \mathrm { H } ^ { + } \right) \times c \left( \mathrm { OH } ^ { - } \right) = 10 ^ { - 12 }$ for [[INLINE_FORMULA_8] when the $K _ { w } = c \left( \mathrm { H } ^ { + } \right) \times c \left( \mathrm { OH } ^ { - } \right) = 10 ^ { - 12 }$ solution of $\mathrm { pH } = \mathrm { a }$ and $\mathrm { pH } = \mathrm { a }$ solution of $c \left( \mathrm { H } ^ { + } \right) = 10 ^ { - \mathrm { a } } \mathrm { mol } / \mathrm { L } , \mathrm { pH } = \mathrm { b }$ are mixed in the same volume with $c \left( \mathrm { H } ^ { + } \right) = 10 ^ { - \mathrm { a } } \mathrm { mol } / \mathrm { L } , \mathrm { pH } = \mathrm { b }$ of $c \left( \mathrm { H } ^ { + } \right) = 10 ^ { - \mathrm { a } } \mathrm { mol } / \mathrm { L } , \mathrm { pH } = \mathrm { b }$ solution of $c \left( \mathrm { H } ^ { + } \right) = 10 ^ { - \mathrm { a } } \mathrm { mol } / \mathrm { L } , \mathrm { pH } = \mathrm { b }$ at the same temperature. INLINE_FORMULA_9]] of sulfuric acid in $c \left( \mathrm { H } ^ { + } \right) = 10 ^ { - \mathrm { a } } \mathrm { mol } / \mathrm { L } , \mathrm { pH } = \mathrm { b }$ of NaOH solution in $c \left( \mathrm { OH } ^ { - } \right) = \frac { 10 ^ { - 12 } } { 10 ^ { - \mathrm { b } } } = 10 ^ { \mathrm { b } - 12 } \mathrm {~mol} / \mathrm { L }$, the two mixed in equal volume of the solution shows neutral, then the acid and base happen to be completely reacted to form a strong acid and strong base salt of sodium sulfate, i.e., $c \left( \mathrm { H } ^ { + } \right) = 10 ^ { - \mathrm { a } } \mathrm { mol } / \mathrm { L } = c \left( \mathrm { OH } ^ { - } \right) = \frac { 10 ^ { - 12 } } { 10 ^ { - \mathrm { b } } } = 10 ^ { \mathrm { b } - 12 } \mathrm {~mol} / \mathrm { L }$, then $c \left( \mathrm { H } ^ { + } \right) = 10 ^ { - \mathrm { a } } \mathrm { mol } / \mathrm { L } = c \left( \mathrm { OH } ^ { - } \right) = \frac { 10 ^ { - 12 } } { 10 ^ { - \mathrm { b } } } = 10 ^ { \mathrm { b } - 12 } \mathrm {~mol} / \mathrm { L }$, the $c \left( \mathrm { H } ^ { + } \right) = 10 ^ { - \mathrm { a } } \mathrm { mol } / \mathrm { L } = c \left( \mathrm { OH } ^ { - } \right) = \frac { 10 ^ { - 12 } } { 10 ^ { - \mathrm { b } } } = 10 ^ { \mathrm { b } - 12 } \mathrm {~mol} / \mathrm { L }$ is the same as $c \left( \mathrm { H } ^ { + } \right) = 10 ^ { - \mathrm { a } } \mathrm { mol } / \mathrm { L } = c \left( \mathrm { OH } ^ { - } \right) = \frac { 10 ^ { - 12 } } { 10 ^ { - \mathrm { b } } } = 10 ^ { \mathrm { b } - 12 } \mathrm {~mol} / \mathrm { L }$. INLINE_FORMULA_13]], so B is correct;
C. In Figure 3, when the values of the horizontal coordinates are equal, the larger the value of the vertical coordinate, the smaller $c \left( \mathrm { X } ^ { - } \right)$ is, and the smaller its solubility product constant is, according to Figure 3, when the horizontal coordinates are the same, $c \left( \mathrm { Cl } ^ { - } \right) > c \left( \mathrm { Br } ^ { - } \right) > c \left( \mathrm { I } ^ { - } \right)$, then $K ( \mathrm { AgCl } ) > K ( \mathrm { AgBr } ) > K ( \mathrm { AgI } )$, and the $0.0100 \mathrm {~mol} / \mathrm { L }$ is used as $0.0100 \mathrm {~mol} / \mathrm { L }$, so B is correct. FORMULA_17]] silver nitrate standard solution, titration of the concentration of both $0.1000 \mathrm {~mol} / \mathrm { LCl } ^ { - } , \mathrm { Br } ^ { - }$ and $\mathrm { I } ^ { - }$ mixed solution, the solubility product constant is small Mr. into the precipitate, so the first precipitation is $\mathrm { I } ^ { - }$, so C is correct;
D. pH is the same monohydric acid diluted the same times, pH value change is larger acidic, hydrochloric acid is a strong acid, acetic acid is a weak acid, then pH value of the same hydrochloric acid and acetic acid diluted the same number of times, the pH value of the larger change is hydrochloric acid, so the I said hydrochloric acid, II said acetic acid; the strength of the conductivity of the solution and the concentration of ions proportional to the $c \left( \mathrm { H } ^ { + } \right) : \mathrm { a } > \mathrm { b } > \mathrm { c }$, the solution conductivity: $\mathrm { a } > \mathrm { b } > \mathrm { c }$, so D error;