What is the molarity of a NaOH solution if 26.55 ml are needed to titrate a .6939g sample of KHP?

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Answer 1 · 2017-06-25T19:37:34

Approx. #0.13*mol*L^-1#............

We need a stoichiometric equation........

#underbrace(C_6H_4(CO_2^(-)K^+)(CO_2H))_"potassium hydrogen phthalate, KHP" + NaOHrarrC_6H_4(CO_2^(-)K^+)(CO_2^(-)Na^+) + H_2O#

There is thus 1:1 stoichiometry........

#"Moles of KHP"=(0.6939*g)/(204.22*g*mol^-1)=3.398xx10^-3*mol.#

And since this molar quantity reacted with #26.55*mL# of titrant...

#"Concentration"=(3.398xx10^-3*mol)/(26.55*mLxx10^-3*L*mL^-1)#

#=0.1280*mol*L^-1#.........