由Ka與溶液濃度計算pH值..

Question

(a) Calculate the pH of a solution that is 0.200 M in NH3 and 0.300 M in NH4Cl. The Ka for NH4+ is 5.70*10-10.
(b) Calculate pH changes that takes place when a 100-mL portion of 0.05 M HCl is added to 400 mL of solution in (a).

英文能力不是很好..所以請用中文回答..Sorry~

Answer 1

Calculate the pH of a solution that is 0.200 M in NH3 and 0.300 M in NH4Cl. The Ka for NH4+ is 5.70x10-10. 解：上述的緩衝溶液其解離方程式如下：　　　　　NH4+ + H2O ----> NH3 + H3O+　解離前：0.3　　　　　　　　 0.2　　　0　解離後：0.3-x　　　　　　　0.2+x　　x　　Ka = [NH3][H3O+]/[NH4+] = 5.70x10-10　　　　Ka = x(0.2+x)/(0.3-x) = 5.70x10-10　　　　　　　　x = 8.55x10-10　　所以 [H+] = 8.55x10-10，其 pH = 9.07。Calculate pH changes that takes place when a 100-mL portion of 0.05 M HCl is added to 400 mL of solution in (a). 解：因所加入的鹽酸會和 NH3 反應，其方程式如下：　　　　HCl + NH3 ----> NH4Cl　　故原有之[NH3]及[NH4+]濃度將會改變為：　　　[NH3] = (0.2x0.4-0.05x0.1)/(0.4+0.1)　　　　　　 = 0.15 M　　　[NH4+]= (0.3x0.4+0.05x0.1)/(0.4+0.1)　　　　　　 = 0.25 M　　　Ka = x(0.15+x)/(0.25-x)= 5.70x10-10 　　　　　　　　x = 9.5x10-10　　所以 [H+] = 9.5x10-10，其 pH = 9.02。