Question
what will be the conc of nitrate ion when 400ml of 0.1 M AgNO3 is mixed with 100 ml of 0.2 m BaBr2
Joshi sir comment
no. of miligm. eq. of AgNO3 = 400*0.1 = 40
no. of miligm. eq. of BaBr2 = 100*0.2*2 = 40
so no of miligm. eq. of Ba(NO3)2 = 40 so no. of milimoles of Ba(NO3)2 = 20
so [NO3-] = 20*2/500 = .08 M