# Thread: calculating ion concentrations2

1. ## calculating ion concentrations2

2.75g of MgO is added to 70mL of 2.40mol/L HNO3. Is the solution that results from the reaction acidic or basic? What is the concentration of the ion that is responsible for the character of the solution?

2. The first step here is to convert the MgO to mol, so we divide by the RMM of MgO.
$\displaystyle \frac {2.75}{16.9994+24.3050} = 0.06658$

Then we divide by the volume in Litres to get molarity.
$\displaystyle \frac {.06658}{.070} = .9511$

In solution, the MgO splits up into $\displaystyle Mg^{2+}$ and $\displaystyle O^{2-}$ ions while the $\displaystyle HNO_3$ splits up into $\displaystyle H^+$ and $\displaystyle NO_3^-$ ions.

Since the $\displaystyle O^{2-}$ ion readily accepts 2 protons, we need to double the molarity to get the number of protons accepted in each Litre of solution. This gives 1.902.

Since the original concentration of $\displaystyle H^+$ was more than 1.902, the solution is acidic, so we are interested in [$\displaystyle H^+$]

This is simply 2.40-1.902 = .4977M = .50M to the correct number of significant figures.