What was the molarity of sodium hydroxide in the waste water?The concentration of sodium hydroxide in waste water from an alumina refinery was found by titrating 20.00 mL aliquots of waste water...
What was the molarity of sodium hydroxide in the waste water?
The concentration of sodium hydroxide in waste water from an alumina refinery was found by titrating 20.00 mL aliquots of waste water against 0.150 M hydrochloric acid, using phenolphthalein as indicator. The average titre of several titrations was 11.40 mL.
molarity (M) refers to the number of moles of a substance in one liter of water.
When you know the concentration (the molarity) of a solution and the volume used you can determine the moles of the substance dissolved in the solution.
In this problem you know the amount of hydrochloric acid (HCl) and the volume of HCl used.
The other information you need is to understand how HCl and sodium hydroxide (NaOH) combine. This is shown in a balanced chemical equation.
HCl + NaOH --> NaCl + HOH
This equation tells you that one mole of HCl will combine with one mole of NaOH to produce one mole of NaCl plus one mole of water.
Going back to your problem:
you use 11.4 mL of 0.15 M HCl.
11.4 mL = 0.0114 L of HCl
0.0114 L * 0.15 M = 0.0017 moles of HCl that reacted with NaOH.
That means the concentration of NaOH is 0.0017 moles in 20 mL.
Now set up a ratio:
0.0017/20 = x/1000 and solve for x
x = 0.0855 M NaOH in the wastewater.