19. If you mix two solutions of equal volume in each option listed below, which option has no buffer action.
A. $0.2 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{HCl}$ 和 $0.2 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{KCl}$
B. $0.02 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{HCl}$ 和 $0.04 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{NH}_{3} \cdot \mathrm{H}_{2} \mathrm{O}$
C. $0.01 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{KH}_{2} \mathrm{PO}_{4}$ 和 $0.2 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{Na}_{2} \mathrm{HPO}_{4}$
D. $0.01 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{NaOH}$ 和 $0.02 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{HAc}$

Step 1 ：Analyze each option:

Step 2 ：A. $0.2 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{HCl}$ and $0.2 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{KCl}$: In this case, we have a strong acid (HCl) and a salt (KCl). There is no weak acid or weak base present, so this option cannot form a buffer solution.

Step 3 ：B. $0.02 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{HCl}$ and $0.04 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{NH}_{3} \cdot \mathrm{H}_{2} \mathrm{O}$: In this case, we have a strong acid (HCl) and a weak base (NH3). When mixed, the HCl will react with NH3 to form NH4+ (ammonium ion), which is the conjugate acid of NH3. This option can form a buffer solution.

Step 4 ：C. $0.01 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{KH}_{2} \mathrm{PO}_{4}$ and $0.2 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{Na}_{2} \mathrm{HPO}_{4}$: In this case, we have a weak acid (KH2PO4) and its conjugate base (Na2HPO4). This option can form a buffer solution.

Step 5 ：D. $0.01 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{NaOH}$ and $0.02 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{HAc}$: In this case, we have a strong base (NaOH) and a weak acid (HAc, acetic acid). When mixed, the NaOH will react with HAc to form Ac- (acetate ion), which is the conjugate base of HAc. This option can form a buffer solution.

Step 6 ：\(\boxed{\text{Final Answer: Option A has no buffer action.}}\)