Question 1 of 8 Map Sapling Learning macmillan leaming Calculate the pH of the s

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Question 1 of 8 Map Sapling Learning macmillan leaming Calculate the pH of the solution after the addition of the following amounts of 0.0695 M HNO3 to a 70.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04 a) 0.00 mL of HNO3 d) 71.5 mL of HNO3 Number Number pH = e) Volume of HNO3 equal to the equivalence point b) 6.79 mL of HNO3 Number Number pH- c) Volume of HNO3 equal to half the equivalence point volume f 80.5 mL of HNO3 Number Number pH- pH= Previous e Che * Answer Next Exit Hint

Explanation / Answer

molarity of aziridine = 0.0750 M = C

a ) 0.00 mL HNO3 added:

pOH = 1/2 [pKb -logC]

= 1/2 [5.96 - log 0.0750]

= 3.54

pH + pOH =14

pH = 10.46

b ) 5.62 mL HNO3 added

millimoles of base aziridine = 0.0750 x 70 = 5.25

millimoles of HNO3 = 6.79 x 0.0695 = 0.472

B + H+ ----------------> BH+

5.25 0.472 0

4.78 0 0.472

pH = pKa + log (4.78 / 0.472)

pOH =8.04 +  log (4.78 / 0.472)

pH = 9.04

c )

at half equivalence point : pOH = pKb

pOH = 5.96

pH = 8.04

d) 71.5mL HNO3 added

millimoles of base aziridine = 0.0750 x 70 = 5.25

millimoles of HNO3 = 4.97

B + H+ ----------------> BH+

5.25 4.97 0

0.281 0 4.97

it is buffer use above formula

pH = 6.79

e ) at equivalece point

only salt remains

salt concetration = 5.25 / 75.54 + 70 = 0.036 M

pH = 7 - 1/2 [pKb + logC]

pH = 7 -1/2 [5.96 + log 0.036]

pH = 4.74

f ) 80.5 mL HNO3

pH = 2.15

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