2233b

计算：

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+的物质的量浓度= 0.2 mol/L

(2)原混合溶液中所含HNO3的物质的量 = 0.01 mol