What will be the concentration of bromide ions on mixing 300ml of 0.1M AgNo3 with 200ml of 0.1M CaBr2 ?
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300 ml of 0.1 M Ag NO3 => 0.1 * 300/1000 = 0.03 moles
200 ml of 0.1 M Ca Br2 => 0.02 moles
2 Ag NO3 (aq) + Ca Br2 (aq) ===> 2 Ag Br + Ca (NO3)2
Hence, 0.03 moles of SilverNitrate with react with 0.015 moles of CaBr2 to form 0.03 moles of Silver Bromide. That is insoluble in water of acids.
Then 0.005 moles of Ca Br2 in the solution are left. If they dissociate into ions. That means 0.01 moles of Br^- ions.
concentration = 0.01 * 1000ml / (300 + 200) ml = 0.02 M
200 ml of 0.1 M Ca Br2 => 0.02 moles
2 Ag NO3 (aq) + Ca Br2 (aq) ===> 2 Ag Br + Ca (NO3)2
Hence, 0.03 moles of SilverNitrate with react with 0.015 moles of CaBr2 to form 0.03 moles of Silver Bromide. That is insoluble in water of acids.
Then 0.005 moles of Ca Br2 in the solution are left. If they dissociate into ions. That means 0.01 moles of Br^- ions.
concentration = 0.01 * 1000ml / (300 + 200) ml = 0.02 M
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