The traditional quantitative chemistry experiment is limited by measurement methods, with narrow research scope and low accuracy. . DIS digital information system (composed of sensors, data,

(1)CH 3 COOH is not completely ionized, but HCl is completely ionized, which makes c(H+) in 0. 1mol/L CH 3 COOH solution smaller than c(H+) in HCl solution with the same concentration, and the pH is larger. Hydrochloric acid reacts with NaOH to produce sodium chloride and water, but sodium chloride does not hydrolyze, while CH 3 COOH reacts with NaOH to produce sodium acetate and water, and sodium acetate hydrolyzes into alkalinity, so it reacts with the same amount of hydrochloric acid and acetic acid, and the solution is neutral after the reaction. The former consumes more NaOH than the latter, so the answer is: CH 3 COOH is not completely ionized, and HCl is completely ionized, so that c(H+) in the 0. 1mol/L CH 3 COOH solution.

(2)① Measure 20.00mL sample of acetic acid solution with acid burette; There is almost no distilled water in the conical flask, and the amount of substances in the liquid to be measured remains unchanged, and V (standard) remains unchanged. According to C (standard) = C (standard) ×V (standard) V (to be measured), C (to be measured) remains unchanged.

So, the answer is: acid burette; No;

② Acetic acid reacts with NaOH solution: CH 3 COOH+NaOH=CH 3 COONa+H 2 O, so the answer is: ch3cooh+OH-= ch3coo-+H2O;

(3) Point B in the figure indicates that acetic acid reacts completely with NaOH, and sodium acetate is hydrolyzed to make C (Na+) > C (COO of CH3-), so the answer is: C (Na+) > C (COO of CH3-) > C (OH-) > C (H+);

④c(H + )= 10 -8？ mol/L，c(OH - )= 10？ - 14 10 ? -8 mol/L= 10 -6 mol/L, and according to the conservation of charge, c (na+)+c (h+) = c (COO of ch 3 -)+c(OH-), c (na+)-c (COO of ch 3-) = c (oh-). mol/L=9.9× 10 -7？ Mol/L, so the answer is: 9.9× 10-7.