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?

1 Answer
anor277
Jun 13, 2017

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

Explanation:

#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#.

commons.wikimedia.org

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#.

Related questions
See all questions in Neutralization
Impact of this question
7024 views around the world
You can reuse this answer
Creative Commons License