If 25.21 mL of #NaOH# solution is required to react completely with .550 g #KHP#, what is the molarity of the #NaOH# solution?

1 Answer
anor277
May 14, 2016

Rxn: #C_6H_4(CO_2H)(CO_2^(-)K^(+)) + NaOH(aq) rarr C_6H_4(CO_2^(-)Na^+)(CO_2^(-)K^(+)) + H_2O#

Explanation:

Given the equation, there is 1:1 equivalence between moles of KHP and moles of sodium hydroxide.

#"Moles of KHP"# #=# #(0.550*g)/(204.22*g*mol^-1)# #=# #2.69xx10^-3*mol#

#"Concentration of NaOH(aq) solution:"# #=# #(2.69xx10^-3*mol)/(25.21xx10^-3*L)# #~=# #0.100*mol*L^-1#, but do the calculation for an exact value.

#"KHP"# is an excellen primary standard, because we could also have used it as a base and not an acid, and used it to standardize a hydrochloirc acid solution.

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