Question

Sketch both on the same graph!

Sketch the curve that represents the titration of 30.0 ml of 0.020 M ammonia, NHs (in the flask) with 0.015 M HCl. K of NE-1.8x 10 (Calculate a few key points; starting pH, pH and volume at equivalence potnt, "halfway, after 50 ml Sketch the curve hat represents the titration of 30.0 ml of 0.020 M pyridine, CsHaN (in the flask) with 0.015 M HNO. K& of CsHsN- 1.7x 109 (Calculate a few key points; starting pH. pH and volume at equivalence point "halfway, after 50 m) 14 1 12 10 8 6 2 0 10 50 20 Vol added (ml) 30 40

Answer 1

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:

14 12 lo Ammmo n nth HCl 3 40 50 30 20 10