CHEM 102 1nd Edition Lecture 29Outline of Last Lecture I. Features of Strong Acid-Base pH CurvesII. Calculate of Solution pHIII. StepsOutline of Current Lecture I. Characteristic Features of Weak Acid-Strong BaseII. Calculation of Solution pHCurrent LectureI. Characteristic Feature of Weak Acid-Strong Base- Initial pH depends on weak acid ionization- pH changes gradually until titration approaches equivalence point (buffer formeddetermines solution pH)- At half-neutralization, pH = pkaII. Calculation of Solution pH- Before addition of any NaOH (pH of 100 mL of 0.10 M CH3COOH The CH3COOH is a weak acid, it is ionized slightly. Determine the amount ionized; x = mol/L CH3COOHCH3COOH H2O H3O+ CH3COO-Initial0.10 M ~ 0 M ~ 0 MReaction- x M + x M + x MEquilibrium(0.10 –x) M X M X M Solve for x by using the Ka expressionH 3O+¿¿CH 3 COO−¿¿¿Ka=¿1.8e-5=x2(0.10−x)∨x2(0.10); x=1.3e-3 [H3O+] = 1.3e-3- After addition of 20.0 mL of 0.100 M NaOH to 100.0 mL of 0.100 M CH3COOHThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.CH3COOH NaOH NaCH3COO H2OInitial0.0100 0.0020 0 Reaction- 0.0020 - 0.0020 + 0.0020 Final0.0080 0 0.0020- [HAc] = 0.0080mol HAc0.120 L=0.0667 M- [NAc] and [C2H3O2]- = 0.0020mol NaAc0.120 L=0.0167 M- Use Henderson pH = pKa + log([X-]/[acid]) pH = -log 1.83-5 + log (0.0167/0.0667) pH =
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