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QUESTION

# The titration of an impure sample of KHP found that 36.0 mL 0.100 M NaOH was required to react completely with 0.765 g of sample. What is the percentage of KHP in this sample?

The ##"KHP"## is 96.1 % pure.

The equation for the reaction with ##"NaOH"## is

##"KHC"_8"H"_4"O"_4 + "NaOH" → "KNaC"_8"H"_4"O"_4 + "H"_2"O"## ##color(white)(mll)"KHP"color(white)(mll) + "NaOH" → color(white)(mll)"KNaP" color(white)(ml)+ "H"_2"O"##

Let's start by calculating the moles of ##"NaOH"##.

##"Moles of NaOH" = 0.0360 color(red)(cancel(color(black)("L NaOH"))) × "0.100 mol NaOH"/(1 color(red)(cancel(color(black)("L NaOH")))) = "0.003 60 mol NaOH"##

Now we can calculate the moles of ##"KHP"## neutralized by the ##"NaOH"##.

##"Moles of KHP" = 0.003 60 color(red)(cancel(color(black)("mol NaOH"))) × "1 mol KHP"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "0.003 60 mol KHP"##

And now we can calculate the mass of the ##"KHP"##.

##"Mass" = "0.003 60" color(red)(cancel(color(black)("mol KHP"))) × "204.22 g KHP"/(1 color(red)(cancel(color(black)("mol KHP")))) = "0.7352 g KHP"##

Finally, we calculate the purity of the ##"KHP"##.

##"Purity" = "mass of pure KHP"/"mass of impure KHP" × 100 % = (0.7352 color(red)(cancel(color(black)("g"))))/(0.765 color(red)(cancel(color(black)("g")))) × 100 % = 96.1 %##

The ##"KHP"## is 96.1 % pure.