titration of 25 ml of 0.100 m hcl with 0.100 m naoh

Tro, Chemistry: A Molecular Approach 1Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCHO2(aq) + NaOH(aq) NaCHO2 (aq) + H2O(aq) after equivalence pointadded 30.0 mL NaOH5.0 x 10-4 mol NaOH xsTro, Chemistry: A Molecular Approach 2added 30.0 mL NaOH0.00050 mol NaOH xspH = 11.96added 35.0 mL NaOH0.00100 mol NaOH xspH = 12.22Adding NaOH to HCHO2added 12.5 mL NaOH0.00125 mol HCHO2pH = 3.74 = pKahalf-neutralizationinitial HCHO2 solution0.00250 mol HCHO2pH = 2.37added 5.0 mL NaOH0.00200 mol HCHOpH = 3.14added 10.0 mL NaOH0.00150 mol HCHOpH = 3.56added 15.0 mL NaOH0.00100 mol HCHO2pH = 3.92added 20.0 mL NaOH0.00050 mol HCHO2pH = 4.34added 40.0 mL NaOH0.00150 mol NaOH xspH = 12.36added 25.0 mL NaOHequivalence point0.00250 mol CHO2CHO2init = 0.0500 M OHeq = 1.7 x 10-6pH = 8.23added 50.0 mL NaOH0.00250 mol NaOH xspH = 12.52Tro, Chemistry: A Molecular Approach 3Tro, Chemistry: A Molecular Approach 4Titration of 25.0 mL of 0.100 M HCHO2 with 0.100 M NaOHThe 1st derivative of the curve is maximum at the equivalence pointSince the solutions are equal concentration, the equivalence point is at equal volumespH at equivalence = 8.23Tro, Chemistry: A Molecular Approach 5at pH 3.74, the HCHO2 = CHO2; the acid is half neutralizedhalf-neutralization occurs when pH = pKaTro, Chemistry: A Molecular Approach 6Titrating Weak Acid with a Strong Base the initial pH is that of the weak acid solutioncalculate like a weak acid equilibrium problem e.g., 15.5 and 15.6 before the equivalence point, the solution becomes a buffercalculate mol HAinit and mol Ainit using reaction stoichiometrycalculate pH with Henderson-Hasselbalch using mol HAinit and mol Ainit half-neutralization pH = pKaTro, Chemistry: A Molecular Approach 7Titrating Weak Acid with a Strong Base at the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established mol A = original mole HA calculate the volume of added base like Ex 4.8 Ainit = mol A/total liters calculate like a weak base equilibrium problem e.g., 15.14 beyond equivalence point, the OH is in excess OH = mol MOH xs/total liters H3O+OH=1 x 10-14Tro, Chemistry: A Molecular Approach 8Ex 16.7a A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the volume of KOH at the equivalence pointWrite an equation for the reaction for B with HA. Use Stoichiometry to determine the volume of added BHNO2 + KOH NO2 + H2OTro, Chemistry: A Molecular Approach 9Ex 16.7b A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOHWrite an equation for the reaction for B with HA. Determine the moles of HAbefore & moles of added BMake a stoichiometry table and determine the moles of HA in excess and moles A madeHNO2 + KOH NO2 + H2OHNO2 NO2- OHmols Before 0.00400 0 0mols added - - 0.00100mols After 00.00300 0.00100Tro, Chemistry: A Molecular Approach 10Ex 16.7b A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH.Write an equation for the reaction of HA with H2ODetermine Ka and pKa for HAUse the Henderson-Hasselbalch Equation to determine the pHHNO2 + H2O NO2 + H3O+HNO2 NO2- OHmols Before 0.00400 0 0mols added - - 0.00100mols After 0.00300 0.00100 0Table 15.5 Ka = 4.6 x 10-4Tro, Chemistry: A Molecular Approach 11Ex 16.7b A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence pointWrite an equation for the reaction for B with HA. Determine the moles of HAbefore & moles of added BMake a stoichiometry table and determine the moles of HA in excess and moles A madeHNO2 + KOH NO2 + H2OHNO2 NO2- OHmols Before 0.00400 0 0mols added - - 0.00200mols After 00.00200 0.00200at half-equivalence, moles KOH = mole HNO2Tro, Chemistry: A Molecular Approach 12Ex 16.7b A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point.Write an equation for the reaction of HA with H2ODetermine Ka and pKa for HAUse the Henderson-Hasselbalch Equation to determine the pHHNO2 + H2O NO2 + H3O+HNO2 NO2- OHmols Before 0.00400 0 0mols added - - 0.00200mols After 0.00200 0.00200 0Table 15.5 Ka = 4.6 x 10-4Tro, Chemistry: A Molecular Approach 13Titration Curve of a Weak Base with a Strong AcidTro, Chemistry: A Molecular Approach 14Titration of a Polyprotic Acid if Ka1 Ka2, there will be two equivalence points in the titrationthe closer the Kas are to each other, the less distinguishable the equivalence points aretitration of 25.0 mL of 0.100 M H2SO3 with 0.100 M NaOHTro, Chemistry: A Molecular Approach 15Monitoring pH During a Titration the general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the H3O+ using a probe that specifically measures just H3O+ the endpoint of the titration is reached at the equivalence point in the titration at the inflection point of the titration curve if you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicatorTro, Chemistry: A Molecular Approach 16Monitoring pH During a TitrationTro, Chemistry: A Molecular Approach 17Indicators many dyes change color depending on the pH of the solution these dyes are weak acids, establishing an equilibrium with the H2O and H3O+ in the solutionHInd(aq) + H2O(l) Ind(aq) + H3O+(aq) the color of the solution depends on the relative concentrations of Ind:HInd when Ind:HInd 1, the color will be mix of the colors of Ind and HInd when Ind:HInd 10, the color will be mix of the colors of Ind when Ind:HInd 0.1, the color will be mix of the colors of HInd18PhenolphthaleinTro, Chemistry: A Molecular Approach 19Methyl RedTro, Chemistry: A Molecular Approach 20Monitoring a Titration with an Indicator for most titrations, the titration curve shows a very large change in pH for very small additions of base near the equivalence point an indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pHpKa of HInd pH at equivalence point 21Acid-Base IndicatorsTro, Chemistry: A Molecular Approach 22Solubility Equilibria all ionic compounds dissolve in water to some degree however, many compounds have such low solubility in water that we classify them as insoluble we can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in waterTro, Chemistry: A Molecular Approach 23Solubility Product the equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, Ksp for an ionic solid MnXm, the dissociation reaction is:MnXm(s) nMm+(aq) + mXn(aq) the solubility product would be Ksp = Mm+nXnm for example, the dissociation reaction for PbCl2 isPbCl2(s) Pb2+(aq) + 2 Cl(aq) and its equilibrium constant is Ksp = Pb2+Cl2Tro, Chemistry: A Molecular Approach 24Tro, Chemistry: A Molecular Approach 25Molar Solubility solubility is the amount of solute that will dissolve in a given amount of solution at a particular temperature the molar solubility is the number of moles of solute that will dissolve in a liter of solution the molarity of the dissolved solute in a saturated solution for the general reaction MnXm(s) nMm+(aq) + mXn(aq)Tro, Chemistry: A Molecular Approach 26Ex 16.8 Calculate the molar solubility of PbCl2 in pure water at 25CWrite the dissociation reaction and Ksp expressionCreate an ICE table defining the change in terms of the solubility of the solidPb2+ ClInitial 0 0Change +S +2SEquilibrium S 2SPbCl2(s) Pb2+(aq) + 2 Cl(aq)Ksp = Pb2+Cl2Tro, Chemistry: A Molecular Approach 27Ex 16.8 Calculate the molar solubility of PbCl2 in pure water at 25CSubstitute into the Ksp expressionFind the value of Ksp from Table 16.2, plug into the equation and solve for SPb2+ ClInitial 0 0Change +S +2SEquilibrium S 2SKsp = P