Suppose 49.6 mL of 0.242 M CoCl2 solution is added to 23.8 mL of 0.361 M NiCl2 solution. Calculate the concentration, in moles per liter, of each of the ions present after mixing. Assume that the volumes are additive.

I'm not sure what to do! Thank you!

Convert M and mL to moles . moles = M x L.

Co will be moles/total volume.
Ni will be moles/total volume.
Cl will be moles of each added and the the total/total volume.

Note that Cl in NiCl2 will be 2x Ni and Cl in CoCl2 will be 2x Co. Do that before adding Cl form one to Cl of the other.

To calculate the concentration of each ion, we need to first determine the number of moles of each ion present in the solutions. Then, we can use the balanced chemical equation to find the final concentration after the solutions are mixed.

Here's how you can solve this problem step by step:

Step 1: Calculate the moles of CoCl2 solution added.
Moles of CoCl2 = volume (in liters) × concentration (in moles per liter)
= 49.6 mL × (1 L / 1000 mL) × 0.242 M
= 0.012 mol

Step 2: Calculate the moles of NiCl2 solution added.
Moles of NiCl2 = volume (in liters) × concentration (in moles per liter)
= 23.8 mL × (1 L / 1000 mL) × 0.361 M
= 0.0086 mol

Step 3: Determine the total moles of each ion.
Co2+: In CoCl2, there is 1 Co2+ ion per CoCl2 molecule, so the total moles of Co2+ = moles of CoCl2 added = 0.012 mol
Ni2+: In NiCl2, there is 1 Ni2+ ion per NiCl2 molecule, so the total moles of Ni2+ = moles of NiCl2 added = 0.0086 mol

Step 4: Calculate the final total volume of the mixture.
Volume = volume of CoCl2 + volume of NiCl2
= 49.6 mL + 23.8 mL
= 73.4 mL

Step 5: Calculate the final concentration of each ion.
Concentration = total moles / final volume (in liters)
Co2+: [Co2+] = moles of Co2+ / volume (in liters)
= 0.012 mol / (73.4 mL × (1 L / 1000 mL))
= 0.163 M
Ni2+: [Ni2+] = moles of Ni2+ / volume (in liters)
= 0.0086 mol / (73.4 mL × (1 L / 1000 mL))
= 0.117 M

Hence, the concentration of Co2+ ions is 0.163 M, and the concentration of Ni2+ ions is 0.117 M in the final mixture after the solutions are mixed.