In standardization, 8.77 mL of #NaOH# neutralized 1.522 g of #KHP#. Given the molar mass of #KHP# is 204.22 g/mol, what is the molarity of the #NaOH# solution?

Answer 1

#"Molarity"=0.8500*mol*L^-1#

#KHP#, #"potassium hydrogen phthalate"#, #1,2-C_6H_4CO_2HCO_2^(-)K^(+)#, is the potassium salt of the diacid, #"pthalic acid"#, #1,2-C_6H_4(CO_2H)_2#.

It can be titrated as a base, or here as an acid. It is air stable, non-hygroscopic, and thus useful as a primary standard.

#1,2-C_6H_4CO_2HCO_2^(-)K^(+)+Na^(+)""^(-)OH rarr "1,2-"C_6H_4CO_2^(-)Na^(+)CO_2^(-)K^(+)+H_2O#

And thus moles of #NaOH# #-=# #"KHP"#, and thus............

#[NaOH]=((1.522*g)/(204.22*g*mol^-1))/(8.77xx10^3*L)=0.8500*mol*L^-1#.

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Answer 2

Kai Bowman

To find the molarity of the NaOH solution, first, calculate the number of moles of KHP used using its molar mass. Then, use the stoichiometry of the reaction to find the number of moles of NaOH. Finally, divide the moles of NaOH by the volume of NaOH used in liters to find the molarity.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.