1. Using the pH obstained for solution 1,2,3 and 4 calculate the percentage ioni
ID: 998426 • Letter: 1
Question
1. Using the pH obstained for solution 1,2,3 and 4 calculate the percentage ionization for each of those 4 solutions.2. Explain the difference in percentage ionization of solutions 1 and 3 as compared to 2 and 4.
3. Calculate the pH change between solutions 7 and 8 and the charge between 10 and 11. A) which changed more? Why? B) Use an equation to show how the buffer neutralized acid
4. Calculate the pH change between solutions 7 and 9 and the charge between 10 and 12. A)Which charged more? Why? B) Use an equation to show how the buffer neutrazed base. 5. Discuss your results and how the buffer affects pH charges. DATA Solution pH Solution pH 1. Strong acid (0.10M HCI) 7. Buffer (NH,+NH) 1 8. Buffer + strong acid (Solution 7 + HCI) 2. Weak acid (0.10M HC,H,o) 3. Strong base (0.10M NaOH) 9. Buffer + strong base (Solution 7+ NaOH) 4. Weak base (0.10M NH,) q.0o 10. water 5. Conjugate of a strong weak acid (0.10M NaC,H,O2) S1 aE rong aid63 11. water strong acid 6. Conjugate of a weak base (0.10M NH NO,) BaC ( ON SMNO613 12 water srong base11.so 12. water+ strong base home solution home solution home solution home solution
Explanation / Answer
1. Using the pH obstained for solution 1,2,3 and 4 calculate the percentage ionization for each of those 4 solutions.
Solution:-
for solution (1) Initial concentration of strong acid = 0.10M
pH = 1.39
[H+] = 10-pH = 10-1.39 = 0.0407
percentage ionization = [H+] x 100/(initial concentration)
percentage ionization = .0407 x 100/0.10 = 40.7%
For solution (2)
[H+] = 10-pH = 10-2.99 = 0.00102
initial concentration = 0.10
percentage ionization = 0.00102 x 100/0.10 = 1.02%
for solution (3) pH = 12.62
pOH = 14 - 12.62 = 1,38
[OH-] = 10-1.38 = 0.0417
intial concentration = 0.10
percentage ionization = 0.0417 x 100/0.10 = 41.7%
for solution (4) pH = 11.03
pH = 14 - 11.03 = 2.97
[OH-] = 10-2.97 = 0.00107
initial concentration = 0.10
percentage ionization = 0.00107 x 100/0.10 = 1.07%
2. Explain the difference in percentage ionization of solutions 1 and 3 as compared to 2 and 4
Solution:- percentage ionization for solution 1 and solution 3 are very close (almost same) since both are strong. for solution 3, the percentage ionization is slightly greater than solution 1 and it would be because of little higher value of its dissociation constant.
Likewise, percentage ionization of solution 2 and 4 are very close(almost same) since both are weak. For solution 4, the percentage ionization is slightly greater than solution 2 and it would because of little higher value of its dissociation constant.
3. Calculate the pH change between solutions 7 and 8 and the charge between 10 and 11.
solution:- pH change between solution 7 and solution 8 = 9.41 - 9.31 = 0.10
pH change between solution 10 and 11 = 2.63- 9.00 = -6.37
A) which changed more? Why?
solution:- Change is high for solution 10 and 11 since solution 10 is water which is slightly basic in nature since its pH is 7. When we add strong acid to it then acid and base neutralize rech other but if the acid is in excess then pH drops donw very rapidly.
B) Use an equation to show how the buffer neutralized acid
in number 7, given buffer is of NH3(weak base) and NH4+(conjugate acid). When an acid is added to this buffer then the acid react with NH3 and forms NH4+ and the added H+ (coming from added strong acid) is neutralized.
NH3 + H+ -------> NH4+
4. Calculate the pH change between solutions 7 and 9 and the charge between 10 and 12.
pH change between solution 7 and 9 = 9.46 - 9.41 = 0.05
pH change between solution 10 and 12 = 11.50 - 9.00 = 2.50
A)Which charged more? Why?
Solution:- pH change is more for solution 10 and 12 since solution 10 is not a buffer, it is slightly basic base and when extra base is added to it then there is more OH- in the solution and pH increases.
B) Use an equation to show how the buffer neutrazed base.
Solution:- A base added to the buffer would react with conjugate acid present in the buffer solution and make a weakly ionized acid and so there is very very low change in pH.
OH- + NH4+ ------------> H2O + NH3
5. Discuss your results and how the buffer affects pH charges.
Solution:- We have noticed that if an acid or base is added to a solution then there is a greater change in pH but same time if an strong acid or base is added to a buffer solution then there is very less change in pH since the buffer solution resists a change to its pH. we call it buffer action,
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