Question

Hello, can someone help me solve this, please? Determine the concentration of NO−3 in each aqueous solution. (Assume complete dissociation of each compound.)

0.18 M KNO 0.18 M KNO_3
0.18 M Ca(NO_3)_2
0.18 M Cr(NO_3)_3

Answer 1

`0.18M`
`0.36M`
`0.54M`

Explanation:

Because the question states "complete dissociation" we will not need to worry about buffers or reactions - JUST the total ion count after dissociation. For that, we simply examine the original material for the part that is the ion of interest (`NO_3^-`) and multiply it by the original molar amount.

0.18 M KNO 0.18 M KNO_3
Because NO reaction is stipulated, we will NOT assume any reaction after the dissociation of the KNO. Thus, the only contribution to the ion amount is `KNO_3`, which is `0.18M`

0.18 M Ca(NO_3)_2
In this case we have TWO `NO_3^-` ions per original compound, so the resulting amount of ions will be doubled: `0.36M`

0.18 M Cr(NO_3)_3
In this case we have three `NO_3^-` ions per original compound, so the resulting amount of ions will be tripled: `0.54M`