Parts that the picture cut off: 0.900M KOH Ka2= 2.01x10^(-6) B) Past the equival
ID: 545207 • Letter: P
Question
Parts that the picture cut off:
0.900M KOH
Ka2= 2.01x10^(-6)
B) Past the equivalence point the pH is due to excess conjugate base, A C)Before strong base is added, the pH is due to the ionization of the weak acid, HA D)Before the equivalence point, the pH is due to a mixture of HA and A E) The pH at the equivalence point is basic 7.A 40.00 mL sample of 0.1000 M diprotic malonic acid is titrated with 0.0900 M KO What volume KOH must be added to give a pH of 5.697? Kal-1.42 x 10% and Ka2-2.0 -6 10 A) 66.67 mL B) 44.44 mL C) 22.22 mL D) 88.89 mL E) 1.11 mL Which of the following are correct for Beers Law? Absorbance increases as concentration increases Absorbance decreases as pathlength increases The molar absorptivity value is chromophore and wavelength specific III re to searchExplanation / Answer
pH = 5.697
Ka2 = 2.01 x 10^-6
pKa = -log Ka
pKa = 5.697
millimoles of acid = 40 x 0.1 = 4
millimoles of acid = millimoles of base
millmoles of KOH = 4
for first equivalence point :
volume of KOH = 4 / 0.09 = 44.44 mL
here in this probelm :
pH = pKa2 . this is second half equivalence point.
volume for half equivalence = 44.44 / 2 = 22.22 mL
total volume = 44.44 + 22.22 = 66.67 mL
volume of KOH = 66.67 mL