�ô-�� ���� 4 ��ԡ����������ҧ�ô-��������� �ô-�� (Acid – Base) �Ũҡ����Դ��ԡ����������ū�ʷ��������������� H3O+ ������� �ռ���� [H3O+ ] > [OH-] �������¨֧�����ѵ��繡ô�� pH < 7 ��ҵ�ͧ��÷�Һ������������� pH ����� ��ͧ��Һ��������鹢ͧ�������� ��з�Һ��� Ka �ͧ�����
�óչ����� Ka �ͧ NH4+ = 5.6 X 10-12 ��ҧ�����Ҩ���͡��� Ka ���µç ��͡��� Kb �ͧ NH4OH ���� NH3 = 1.8 X 10-5 �������ö�ӹdz�Ҥ�� Ka �� �ҡ����� Ka.Kb = Kw �ҡ��������� �ѧ����� ; NH4+ + H2O ↔ NH3 + H3O+ ; Ka = 5.6 X 10-12 �ӹdz�� [H3O+ ] �������� pH �ͧ�������´ѧ���
[H3O+]2 = 5.6 X 10-12 CH3COOH + NaOH ↔ CH3COONa + H2O CH3COONa �����ͷ���Դ�ҡ�ô�� + ���� ���ͪ�Դ������������¹�����Ǩ��Դ��ԡ����������ū�� ��������㹢���á���繴ѧ����� CH3COONa → CH3COO- + Na+ �������¹�ŧ㹢�鹵����繡���Դ��ԡ����������ū�� ����ͧ�鹨��Ѻʹ�����Ҥ��繵���Դ��ԡ����������ū�� ��������ѡ������������ ��Ҥ����Ҩҡ�ô�����繵���Դ��ԡ����������ū�� ���ʴ����ѵ������� �ѧ����� �Ũҡ����Դ��ԡ����������ū�ʷ��������������� OH- ������� �ռ���� [OH-] > [H3O+ ] �������¨֧�����ѵ������� ��ҵ�ͧ��÷�Һ pH �ͧ�������� ��ͧ��Һ��������鹢ͧ�������� ��з�Һ��� Kb �ͧ���ô �óչ�����
Kb �ͧ CH3COO- = 5.6 X 10-12 ��ҧ�����Ҩ���͡��� Kb ���µç ��͡��� Ka �ͧ CH3COOH = 1.8 X 10-5 �������ö�ӹdz�Ҥ�� Kb �� �ҡ����� Ka.Kb = Kw �ҡ��������� �ѧ����� ; CH3COO- + H2O ↔ CH3COOH + OH- ; Kb = 5.6 X 10-12 �ӹdz�� [OH- ] �������� pOH ��� pH �ͧ�������´ѧ���
[OH-]2 = 5.6 X 10-12
�ô�� ; HCOOH
↔ HCOO- + H+ ; Ka = 1.77 x 10-4 �������·������� pH < 7 ��ҵ�ͧ��÷�Һ����� pH ����� ��ͧ��Һ��������鹢ͧ�������� (�������� = 1M) ��е�ͧ��Һ��� Ka �ͧ���ô�ͧ���� 㹷������� Ka �ͧ NH4+ ������͡��Ҵѧ��������µç ��͡��� Kb �ͧ NH4OH = 1.8 x 10-5 ���� Ka �ͧ NH4+ ��ҡ����� Ka.Kb = Kw �ѧ��� ; �ʴ�������������Ŵѧ��� ; NH4+ + H2O ↔ NH3 + H3O+ ; Ka = 5.6 x 10-10 �ӹdz�� [H3O+ ] �������� pH �ͧ�������´ѧ��� 4.2 ��� Kb>Ka ���ͷ����ҡ��ԡ����Ҩ������� (weak base)
������ pH >7 �� �������¨������� pH > 7 ��ҵ�ͧ��÷�Һ������������� pH ����� ��ͧ��Һ��������鹢ͧ�������� (�������� = 1M) ��е�ͧ��Һ��� Kb �ͧ����ʢͧ�ô HCN 㹷������� Kb �ͧ CN- ������͡��Ҵѧ��������µç ������ҡ����� Ka.Kb = Kw �ʴ�������������Ŵѧ��� ; CN- + H2O ↔ HCN + OH- ; Kb = 3.08 x 10-5 �ӹdz�� [OH- ] �������� pOH ��������¹�� pH �ͧ�������� �ѧ��� Kb = [HCN][OH-] / [CN-] [OH-]2
= 3.08 x 10-5 = 14 - 2.26 Ẻ�֡�Ѵ 1. Consider the reaction: CH3NH2(aq) + H2O(l) ↔ CH3NH3+(aq) + OH-(aq) where Kb = 4.4 X 10-4. To a solution formed from the addition of 2.0 mol
CH3NH2 to 1.0 L of H2O is added 1.0 mol of KOH (A) 3.2 x 10-2 M (B) 2.2 x 10-4 M (C) 2.0 x 10-3 M (D) 8.8 x 10-4 M (E) None of these 2. The Kb value of the oxalate ion,
C2O42-, is 1.85 X 10-10. Is a solution of K2C2O4 acidic, basic, or neutral? Explain. 3. Consider the following compounds and suppose that 0.5 M solutions are prepared of each: NaI,
KF, 4. Complete and balance the following neutralization / acid-base reactions: (����¹��д��������������ó�) 6. Given a salt, predict an acid-base pair which would produce the salt: a) Al2(SO4)3 c) CaCl2 b) NH4F d) KBr 7. If in the reaction shown above, 25.0 mL of 0.50 Molar HCl is used to neutralize
17.0 mL of NaOH, what is the 8. If in the reaction shown above, 23.5 mL of 0.55 Molar NaOH is used to neutralize 33.0 mL of HCl, what is the molarity of the HCl?
(�������� NaOH ��������� 0.55M �ӹǹ 23.5 mL �ӻ�ԡ����ҾʹաѺ�������� HCl 9. What is the concentration of a sodium hydroxide solution when 30.0mL of 0.50M hydrochloric acid are needed 10. What is the concentration of acetic acid in vinegar when 32.5mL of 0.56M sodium hydroxide is needed to neutralize 15.0mL of the vinegar? (��������������δ�͡䫴��������� 0.56M �ӹǹ 32.5mL �� 11. List the conjugate acids of: (that means the following would behave like bases…) (���Ԩ�ó���Ҩҡ��Ҥ��� a) H2O b) NH3 c) OH- d) HPO42- 12. List the conjugate bases of: (that means the following would behave like acids…)(���Ԩ�ó���Ҩҡ��Ҥ��� a) H2O b) HS- c) HCl d) NH4+ 13.
Determine the pH of 0.05 M CH3COONa. (Ka = 1.8 x 10-5 for acetic acid) (�������� CH3COONa ��������� 0.05 M 14. Determine the pH of 0.05 M NH4NO3. (Kb= 1.8 x 10-5 for ammonia) (�������� NH4NO3 ��������� 0.05M �� pH ����� 15. Calculate the pH of a solution prepared by dissolving 10.5 g of NaF in 500.0 mL of water. 16..
Aspirin is acetylsalicylic acid, a monoprotic acid whose Ka value is 3.27 X 10-4. Does a solution of the sodium 17. Calculate the pH of a 0.10 M NaC2H3O2 solution. Ka for HC2H3O2 = 1.8 x 10-5. (���ӹdz��� pH �ͧ�������� 18. 3.2 M sulfuric acid is used to neutralize 75.0 mL of 2.1 M potassium hydroxide. 19. The base anhydride calcium oxide, CaO can be added to
water to make a solution of calcium hydroxide. If �ٻ�Ҿ�������Ǣ�ͧ |