www2233b
計算：
1 [H+] =[OH-] = 50x4/1000 = 0.2 (mol/L) ；n = CV = 0.2x100/1000 = 0.02 (mol)
2m(H2SO4) = (1.165/233)X98 = 0.49 (g) ；
n(H2SO4) = 0.49/98 = 0.005 mol ; n (H+)= 2x0.005 = 0.01 mol
3n(HNO3) = n - n (H+) = 0.02 (mol) - 0.01 mol = 0.01 mol
答：(1)原混合溶液中H+的物質的量濃度= 0.2 mol/L
(2)原混合溶液中所含HNO3的物質的量 = 0.01 mol