100ml溶液中,有0.01mol-L的h2so4,0.04mol-L的hno3,加入1.92gCU,加热充分反映后,求溶液中铜离子浓度.
问题描述：
100ml溶液中,有0.01mol-L的h2so4,0.04mol-L的hno3,加入1.92gCU,加热充分反映后,求溶液中铜离子浓度.
答
n(Cu) = 1.92/64 = 0.03moln(NO3-) = 0.04*0.1 = 0.004moln(H+) = 0.01*0.1*2+0.04*0.1 = 0.006molCu只能和HNO3反应：3Cu + 8H+ + 2NO3- == 3Cu2+ + 2NO↑ + 4H2O反应的H+：NO3- = 8:2 = 4:1所以H+少量,要按照H+的量...