In this example problem, we calculate the concentration of ions when adding together two solutions. Our problem reads, "If 50 milliliters of 0.050 molar aluminum sulfate is added to 150 milliliters of 0.080 molar sodium sulfate, what is the sulfate ion concentration in the final solution?" If we want to calculate the molarity of sulfate ions, our equation is moles of sulfate divided by the liters of solution. First, we can find the total volume of the solution. We'll want to convert our aluminum sulfate solution volume from milliliters to liters, and we'll do the same with the sodium sulfate solution volume. Assuming the solution volumes are additive, our total solution volume should be 0.2 liters. Next, we can find the total moles of sulfate. We'll use dimensional analysis to convert from 50 milliliters of aluminum sulfate solution to moles of sulfate. First, we'll convert from milliliters to liters using our prefix multiplier for milli. Then we can use the molarity of aluminum sulfate to convert from liters of solution to moles of aluminum sulfate. Then we use the chemical formula to convert from moles of aluminum sulfate to moles of just sulfate: 1 mole of aluminum sulfate contains 3 moles of sulfate ions. Completing the calculation, we have 0.0075 moles of sulfate ions from the aluminum sulfate solution. Next, we can convert 150 milliliters of sodium sulfate solution to moles of sulfate ions. We can convert from milliliters to liters first using our prefix multiplier. Then we use the molarity of sodium sulfate to convert from liters of solution to moles of sodium sulfate. Then we use the chemical formula to convert from moles of sodium sulfate to moles of just sulfate. We have one mole of sulfate for every mole of sodium sulfate. Completing the calculation, we get 0.012 moles of sulfate in the sodium sulfate solution. We can add these together to find the total moles of sulfate as we plug it into our molarity equation at the top. Completing our calculation, our final solution should be 0.098 molar sulfate ions.