Sapling Learning The pK, values for the dibasic base B are pKs 210 and pKe 771.

Question

Sapling Learning The pK, values for the dibasic base B are pKs 210 and pKe 771. caalole th pHat each of the following points in he tratio, of 500 mL of a 0 90 M B(ag)with 090 M Haag. Number (a) before addition of any HC 12.93 Number For part (b).consider that the concentrations of the acid and base are equal. Thus, the first equavalene point o ours when the volumes are eual (50.0 ml of each). 25.0 mL is half-way to the finst equivalence point, so pOHpk (b) afer addition of 25.0 ml. of HC1 12.9 Number (c) after addition of 50 0 mL of HC6.12 Number (d) after addition of 75.0 mL of HC659 Nuncer (e) after addition of 100.0 ml of HC3.32 acBook Pro

Explanation / Answer

b)

mmol of base initially =MV = 0.9*50 = 45 mmol

mmol of acid added = MV = 0.9*25 = 22.5 mmol

then

mmol of base reacted = 45 - 22.5 = 22.5

mmol of conjguate formed = 0+ 22.5 = 22.5

this is half point!

in the half point

pOH = pKb + log(BH+/B)

since  BH+/B = 22.5/22.5 = 1

log(1) = 0

pOH = pKb1

pOH = 2.10

pH = 14-pOH = 14-2.10 = 11.90

c)

mmol of base initially =MV = 0.9*50 = 45 mmol

mmol of acid added = MV = 50*0.9 = 45

then

mmol of base reacted = 45 - 45 = 0

mmol of acid lef t= 45-45

this is first equivalence point

since this is diprotic base

then

pOH = 1/2*(pKb1 + pKb2)

pOH = 1/2*(2.10+7.71)

pOH = 4.905

pH = 14-pOH = 14-4.905 = 9.095

d)

mmol of base initially =MV = 0.9*50 = 45 mmol

mmol of acid added = MV = 75*0.9 = 67.5

mmol of 1st --> 45-45 = 0

mmol of acid left = 67.5-45 = 22.5

now..

mmol of 2nnd base --> 45 - 22.5 = 22.5

mmol of conjguate formed = 22.5

this is 2nd half euqivalence point

in this point

pOH = pKb2 + log(HBH++ / BH+)

pOH = 7.71 + log(22.5/22.5)

pOH = 7.71

pH = 14-7.71 = 6.29

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