a)
Given 30 mL of 0.020 M ammonia
Number of moles of ammonia = 0.02 M * 0.03 L = 0.0006 moles
Kb of ammonia = 1.8 x 10-4
Concentration of OH- = [(Concentration of ammonia)*(Kb)]1/2
Concentration of OH- = [(0.02)*1.8 x 10-4] = 1.89 x 10-3 M
pOH = -log (Concentration of OH-) = -log(1.89 x 10-3 )
pOH = 2.72
Initial pH = 11.27
For volume at equivalence point,
moles of ammonia should be equal to moles of HCl
So, moles of HCl = 0.0006 = Molarity of HCl * Volume of HCl
Given, molarity of HCl = 0.015 M
Volume of HCl at equivalence point = 0.0006/0.015 = 0.04 L or 40 mL
pH at equivalence point:
As NH4+ ions are formed i.e. 0.0006 moles
Total volume at equivalence point = 70 mL or 0.07 L
Concentration of NH4+ ions = 0.0006/0.07 = 0.00857 M
Now, equation involved is:
NH4+ + H2O <------> NH3 + H3O+
Initial 0.00857 0 0
At equi 0.00857 - x x x
Ka = x2 / (0.00857-x)
Also, Ka = Kw/Kb = 1 x 10-14 / 1.8 x 10-4
Ka = 5.56 x 10-11
5.56 x 10-11 = x2 / (0.00857-x)
taking 0.00857-x = 0.00857 (as x is small)
x = (0.00857*5.56 x 10-11 )1/2 = 0.069 x 10-5 or 6.9 x 10-7
This is concentration of H+.
pH = -log( 6.9 x 10-7) = 6.16
So, pH at equivalence point = 6.16
b)
For pyridine
Given 30 mL of 0.020 M pyridine
Number of moles of pyridine = 0.02 M * 0.03 L = 0.0006 moles
Kb of pyridine = 1.7 x 10-9
Concentration of OH- = [(Concentration of ammonia)*(Kb)]1/2
Concentration of OH- = [(0.02)*1.7 x 10-9] = 5.8 x 10-6 M
pOH = -log (Concentration of OH-) = -log(5.8 x 10-6 )
pOH = 5.25
Initial pH = 8.75
For volume at equivalence point,
moles of ammonia should be equal to moles of HNO3
So, moles of HNO3 = 0.0006 = Molarity of HNO3 * Volume of HNO3
Given, molarity of HNO3 = 0.015 M
Volume of HCl at equivalence point = 0.0006/0.015 = 0.04 L or 40 mL
pH at equivalence point:
As pyridinium ions are formed i.e. 0.0006 moles
Total volume at equivalence point = 70 mL or 0.07 L
Concentration of pyridinium ions = 0.0006/0.07 = 0.00857 M
Now, equation involved is:
pyridinium + H2O <------> pyridine + H3O+
Initial 0.00857 0 0
At equi 0.00857 - x x x
Ka = x2 / (0.00857-x)
Also, Ka = Kw/Kb = 1 x 10-14 / 1.7 x 10-9
Ka = 5.88 x 10-5
5.88 x 10-5 = x2 / (0.00857-x)
taking 0.00857-x = 0.00857 (as x is small)
x = (0.00857*5.88 x 10-5 )1/2 = 0.71 x 10-3 or 7.1 x 10-4
This is concentration of H+.
pH = -log( 7.1 x 10-4) = 3.15
So, pH at equivalence point = 3.15.
Graph:
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