Question

The following solutions are mixed together in a single beaker, calculate the total concentration of ions in the final solution?

250 mL of 0.1 M sodium chloride

100 mL of 0.1 M calcium nitrate

150 mL of 0.1 M sodium nitrate

Answer 1

1 mol of NaCl has 2 moles of ions

So,

mol of ions from NaCl = 2*M(NaCl)*V(NaCl)

= 2*0.1*250 mL

= 50 mmol

1 mol of Ca(NO3)2 has 3 moles of ions

So,

mol of ions from Ca(NO3)2 = 3*M(Ca(NO3)2)*V(Ca(NO3)2)

= 3*0.1*100 mL

= 30 mmol

1 mol of NaNO3 has 2 moles of ions

So,

mol of ions from NaNO3 = 1*M(NaNO3)*V(NaNO3)

= 2*0.1*150 mL

= 30 mmol

Total number of mol of ions = 50 mmol + 30 mmol + 30 mmol

= 110 mmol

total volume = 250 mL + 100 mL + 150 mL

= 500 mL

Now use:

concentration of ions = Total number of mol of ions / total volume

= 110 mmol / 500 mL

= 0.22 M

Answer: 0.22 M