Caculate the titration curve for the titration of 20.0 mL of a 0.05 M solution of HCl. The titrant is a 0.01 M solution of NaOH. Why is it necessary to experimentally determine the molarity of the NaOH solution used in the titrations in this(volumetric analysis-acid/base titration) experime...
普通化学原理acid-base titration-ppt
§5.1TitrationCurve§5.2Acid-baseIndicators§5.3TitrationError§5.4Application §5.1Titrationcurve §5.1.1StrongAcidVersusStrongBase Supposingthat20.00ml0.1000mol/LHClistitratedwith0.1000mol/LNaOH.(1)Beforethetitration 0.1000mol/LHCl;pH=1(2)When18.00mlofNaOHsolutionisadded [H+]=0.1000(...
HCl+NaOH = Na++Cl-+H2O pH 5.0-9.0: HOAc+NaOH = -OAc+Na++H2O pH 9.0-11.5: NH4Cl+NaOH = Na++Cl-+NH3+H2O R1=Q1*C/1, C=1/z (HCl in mol/L) R2=Q2*C/1, C=1/z (HOAc in mol/L) R3=Q3*C/1, C=1/z (NH4Cl in mol/L) R4=BETAHNV1, unit: mmol/(L*pH) R5=BETAHN...
Caculate the titration curve for the titration of 20.0 mL of a 0.05 M solution of HCl. The titrant is a 0.01 M solution of NaOH. The end point of a titration is not the same as the equivalence point of a titration. Differentiate between these two...
2[CO3 2− ]x +[NH3] + +[A[−H]C+O[3O−H]x−+] −3[[PHO+4]3−]x + 2[ H PO4 2− ]x + [ H2 PO4 − ]x (Vs + Vx) (1) where: Ve = volume of titrant (strong acid) required to the equivalence point (L); Vx = volume of titrant added to yield ...