Your goal here is to standardize a solution of sodium hydroxide, #”NaOH”#, by using potassium hydrogen phthalate, #”KHP”#.
Your starting point here is the balanced chemical equation for this neutralization reaction
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#”KHP”_text((aq]) + “NaOH”_text((aq]) -> “KNaP”_text((aq]) + “H”_2″O”_text((l])#
The important thing to notice here is that you have a #1:1# mole ratio between the two reactants. This tells you that at you can reach the equivalence point by reacting equal number of moles of #”KHP”# and of #”NaOH”#.
You start with #”0.5100 g”# of #”KHP”#. To get the molar amount of acid used for the experiment, use its molar mass
#0.5100color(red)(cancel(color(black)(“g”))) * overbrace(“1 mole KHP”/(204.22color(red)(cancel(color(black)(“g”)))))^(color(purple)(“molar mass of KHP”)) = “0.0024973 moles KHP”#
So, you know that at equivalence point, the reaction will consume #0.0024973# moles of #”KHP”# and #0.0024973# moles of #”NaOH”#, since that’s what the #1:1# mole ratio tells you.
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As you know, molarity is defined as moles of solute per liters of solution.
#color(blue)(|bar(ul(color(white)(a/a)c = n_”solute”/V_”solution”color(white)(a/a)|)))#
You know how many moles of sodium hydroxide were needed to reach the equivalence point, and the volume of sodium hydroxide solution that delivered that many moles to the reaction.
This means that you can calculate its actual molarity by dividing the two – do not forget to convert the volume from milliliters to liters by using the conversion factor
#”1 L” = 10^3″mL”#
In your case, you will have
#[“NaOH”] = “0.0024973 moles”/(31.70 * 10^(-3)”L”) = color(green)(|bar(ul(color(white)(a/a)”0.07878 M”color(white)(a/a)|)))#
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As you can see, the actual molarity of the solution you prepared is relatively small compared with your target of #”0.100 M”#.
You can calculate the percent error by using the formula
#color(blue)(“% error” = (|”approximate value” – “exact value”|)/”exact value” xx 100)#
In your case, you will get
#”% error” = (|0.07878 – 0.100|)/0.100 xx 100 = 21.22%#
This is characteristic of a major error in your experiment.
My guess is that you allowed too much sodium hydroxide to react with the acid, which would cause the molarity of the solution to appear to be smaller than in reality.
If you’re using phenolphthalein as your indicator, an excess of sodium hydroxide would cause the solution to be a brighter shade of pink than it should be at equivalence point.
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