A solution of sodium hydroxide is standardized against potassium hydrogen phthalate. From the following data, calculate the molarity of NaOH solution? (See details below)

Mass of KHP used: 0.4536 g
Buret reading before titration: 0.23 ml
Buret reading

How many liters of the standardized NaOH will be needed to neutralize a water solution with 0.1453 g Oxalic acid dissolved?

1 Answer
Mar 9, 2017

Answer:

We need (i) a stoichiometric equation........

Explanation:

We need (i) a stoichiometric equation:

#"1,2-C"_6"H"_4"(CO"_2"H)(CO"_2^(-)"K"^(+)")"+"NaOH"(aq) rarr "1,2-C"_6"H"_4("CO"_2^(-)"Na"^(+))("CO"_2^(-)"K"^(+))+"H"_2"O"#

And (ii) stoichiometric quantities of #"KHP"#

#=(0.4536*g)/(204.22*g*mol^-1)=2.221xx10^-3*mol#

Of course, you have not quoted a volume of titrant; i.e. the end reading on the burette. And thus we cannot assess the molarity of the sodium hydroxide solution. Once such a volume is reported, we make the following calculation:

#(2.221xx10^-3*mol)/("Volume of titrant " cancel(mL))xx10^3*cancel(mL)*L^-1=??*mol*L^-1#.

A stoichiometrically balanced equation is absolutely basic to such acid/base titrations.

For the next reaction with the standardized titrant, we follow the stoichiometric equation:

#HO(O=C-C=O)OH(aq) + 2NaOH(aq) rarr Na^(+)""^(-)O(O=C-C=O)O^(-)Na^(+) + 2H_2O(l)#