A titration is performed on 50.0 mL of 0.20 M ammonia with 0.10 M HCI. a. What i

Question

A titration is performed on 50.0 mL of 0.20 M ammonia with 0.10 M HCI. a. What is the pH at the starting point? b. What volume of HCl solution is required to reach the equivalence point? c. What would be the pH after every 20.0 mL of HCl were added up to and including the equivalence point? d. What is the pH at the midpoint of the titration? e. Determine the pH after 20.0 additional mL of HCl has been added beyond the equivalence point. f Sketch the pH curve (or graph using Excel, etc). Label the buffer zone, the midpoint, and the equivalence point.

Explanation / Answer

Solution:-

(a) In this titration NH3, a weak base is titrated against a strong acid, HCl. Starting point pH means the pH when no acid is added to NH3. Since NH3 is a weak base so we will use ICE table to calculate the pH. Kb for NH3 is 1.8 x10-5. Concentration of it is given as 0.20 M. Initially the products will be zero. Let's say the change in concentration is X. The ice table will be as follows...

NH3(aq) + H2O(l) <---------> NH4+(aq) + OH-(aq)

I 0.20 0 0

C -X +X +X

E (0.20 - X) X X

Kb = [NH4+(aq)] [OH-(aq)]/[ NH3(aq)]

1.8 x10-5 = [(X)2]/[0.20 - X]

since the value of Kb is very low, so the value of X will also be low and hence the X could be neglected for the denominator.

1.8 x10-5 = [(X)2]/[0.20]

(X)2 = 1.8 x10-5 x 0.20

(X)2 = 3.6 x 10-6

taking square root to both sides..

X = 1.897 x 10-3

From ICE table, [OH-] = X

So, [OH-] = 1.897 x 10-3

pOH = - log  [OH-]

pOH = - log (1.897 x 10-3)

pOH = 2.72

pH = 14 - 2.72

pH = 11.28

(b) Equivalece point is the point at which the moles of base are equal to the moles of acid required to neutralize it.

NH3 and HCl react in 1:1 mol ratio which is clear from the below balanced equation.

NH3  + HCl ----> NH4Cl

50 ml of 0.20 M NH3 are taken. So, the moles of NH3 = 50 ml x (1L/1000 ml) x (0.20 mol/L) = 0.01 mol

Since, there is 1:1 mol ratio beteen base and acid, the moles of HCl would also be 0.01.

Molarity of HCl used is given as 0.10 M.

So, volume of HCl used = 0.01 mol x (1L/0.10 mol) x (1000 ml/1L) = 100 ml

(c) As per the given instructions, 20 ml of HCl were added each time till the equivalence point is reached. So, we will calculate the pH for each 20 ml of HCl added to the base.

(i) When 20 ml of HCl are added then, moles of HCl added would be = 20 ml x (1L/1000ml) x (0.10 mol/L) = 0.002 mol

Initial moles of NH3 = 50 ml x (1L/1000 ml) x (0.20 mol/L) = 0.01 mol

excess moles of NH3 = 0.01 - 0.002 = 0.008 mol

moles of NH4+ formed in the reaction (NH3  + HCl ----> NH4Cl) would be equal to the moles of HCl added that is 0.002 mol since HCl is limiting reactant. Here we have a buffer solution so we could use the Handerson equation to calculate the pH.

Total volume of the solution will be 50 ml + 20 ml = 70 ml = 0.07 L

so, [NH3] = 0.008 mol/0.07L = 0.114 M

[NH4+] = 0.002 mol/0.07L = 0.0286 M

pH = Pka + log(base/acid)

Pka could be calculated from the given Kb as..

Pkb = - log(Kb)

Pkb = - log(1.8 x 10-5) = 4.74

Pka = 14 - Pkb

Pka = 14 - 4.74 = 9.26

Plugging in the values in Handerson equation...

pH = 9.26 + log(0.114/0.0286)

pH = 9.26 + 0.60

pH = 9.86

(ii) When 20 more ml that is total 40 ml of HCl are added..

moles of HCl added = 40 ml x (1l/1000ml) x (0.10 mol/L) = 0.004 mol

Moles of NH3 as calculated before = 0.01 mol

excess moles of NH3 = 0.01 - 0.004 = 0.006

moles of NH4+ formed = 0.004 mol

Total volume = 50 ml + 40 ml = 90 ml = 0.09 L

[NH3] = 0.006 mol/0.09L = 0.0667 M

[NH4+] = 0.004 mol/0.09 L = 0.0444 M

pH = 9.26 + log(0.0667/0.0444)

pH = 9.26 + 0.18

pH = 9.44

(iii) When again 20 ml of HCl that is 60 ml of HCl in total are added.

moles of HCl added = 60 ml x (1L/1000 ml) x (0.10 mol/L) = 0.006 mol

moles of NH3 = 0.01 mol

excess moles of NH3 = 0.01 - 0.006 = 0.004 mol

moles of NH4+ formed = 0.006 mol

total volume = 50 ml + 60 ml = 110 ml = 0.11 L

[NH3] = 0.004 mol/0.11L = 0.0364 M

[NH4+] = 0.006 mol/0.11L = 0.0545 M

pH = 9.26 + log(0.0364/0.0545)

pH = 9.26 - 0.18

pH = 9.08

(iv) When 20 ml of HCl are again added that is 80 ml of HCl in total are added.

moles of HCl = 80 ml x (1L/1000ml) x (0.10 mol/L) = 0.008 mol

moles of NH3 = 0.01 mol

excess moles of NH3 = 0.01 - 0.008 = 0.002 mol

moles of NH4+ formed = 0.008 mol

Total volume = 50 ml + 80 ml = 130 ml = 0.13L

[NH3] = 0.002 mol/0.13L = 0.0154 M

[NH4+] = 0.008 mol/0.13L = 0.0615 M

pH = 9.26 + log(0.0154/0.0615)

pH = 9.26 - 0.60

pH = 8.66

(v) When 20 ml of HCl are again added means 100 ml of HCl in total, the equivalence point is reached and at this point we have equal moles of these. So, we will have only 0.01 mol of NH4+ at this point and total volume will be = 50 ml + 100 ml = 150 ml = 0.15 L

[NH4+] = 0.01 mol/0.15 L = 0.0667 M

the equation will be....

NH4+   + H2O <----> NH3 + H3O+

I 0.0667 0 0

C -X +X +X

E (0.0667 - X) X X

Ka = (X)2/(0.0667 - X)

Ka could be calculated using Pka.

Ka = 10-Pka

Ka = 5.5 x 10-10

5.5 x 10-10 = (X)2/(0.0667 - X)

since the Ka value is very low so again the X on the bottom could be neglected.

5.5 x 10-10 = (X)2/(0.0667)

(X)2 = 3.67 x 10-11

taking square root to both sides..

X = 6.06 x 10-6

pH = - log(H3O+]

pH = - log(6.06 x 10-6)

pH = 5.22

(d) Midpoint of the half equivalence point is the point at which exactly half of the equivalence point volume of HCl is added. From previous calculations, 100 ml of HCl are added at the equivalence point. So, the volume of HCl added at mid point is 50 ml.

moles of HCl added = 50 ml x (1L/1000ml) x (0.10mol/L) = 0.005 mol

moles of NH3 = 0.01 mol

Excess moles of NH3 = 0.01 - 0.005 = 0.005 mol

moles of NH4+ formed = 0.005 mol

Total volume = 50 ml + 50 ml = 100 ml = 0.1 L

[NH3] = 0.005 mol/0.1 L = 0.05 M

[NH4+] = 0.005 mol/0.1 L = 0.05 M

pH = 9.26 + log(0.05/0.05)

pH = 9.26

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