posted by Fai .
5ml of 8N Hn03, 4.8ml of 5N hcl and a certain volume of 17M h2s04 are mixed together and made up to 2 liter. 30ml of this acid mixture exactly neutralizes 42.9ml of na2c03, 10 h20 in 100ml of water
.Calculate the amount of sulphate ion in gm present in solution
Who help step by step clearly explain
This can't be done with the information provided. You need a N or M of Na2CO3.10H2O.