# 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?

Mar 9, 2017

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

#### Explanation:

We need (i) a stoichiometric equation:

$\text{1,2-C"_6"H"_4"(CO"_2"H)(CO"_2^(-)"K"^(+)")"+"NaOH"(aq) rarr "1,2-C"_6"H"_4("CO"_2^(-)"Na"^(+))("CO"_2^(-)"K"^(+))+"H"_2"O}$

And (ii) stoichiometric quantities of $\text{KHP}$

$= \frac{0.4536 \cdot g}{204.22 \cdot g \cdot m o {l}^{-} 1} = 2.221 \times {10}^{-} 3 \cdot m o l$

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)