I have attempted an answer to the question below. Please let me know if my reasoning is correct. Thanks.
Q) You have a 1L solution of barium chromate, where the concentration of chromate ions is 110nM. How many mols of barium chloride should be added in order to reduce the concentration of the chromate ions to 12nM?My attempt
Equation for barium chromate in solution:
When we add barium chloride, the balanced equation will be:
We are told that [CrO42-
] = 110nM, and we want to reduce it to 12nM. We have 1L, so that means we want to go from 110 mols down to 12 mols (mols = M×volume). This is a difference of 98 mols. From the equations above, 1 mol of barium chloride eliminates 1 mol of chromate ions. Therefore the mols of barium chloride required to consume 98 mols of chromate ions are 98 mols of barium chloride. So, my answer is 98 mols of BaCl2
Are my steps of reasoning correct? I don't know what the correct answer is.