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

Mar 10, 2017

$M \left(N a O H\right) = 0 , 0688 \frac{m o l}{L}$
$V \left(N a O H\right) = 0 , 0469 L$

#### Explanation:

For a titration is necessary that the moles of NaOH are equivalent to the mole of KHP (that have a MM of 204,22g/mol).

$m o l K H P = \frac{0 , 436 g}{204 , 22 \left(\frac{g}{m o l}\right)} = 0 , 00213 m o l$.

V of NaOH used =(31,26-0,23) = 31,03 mL = 0,03103 L

$m o l N a O H = M V = M 0 , 03103 L$
but the mol of the two substance are the same therefore
Molarity of NaOH = (mol KHP)/ (V NaOH used for titration): $M \left(N a O H\right) = \frac{0 , 00213 m o l}{0 , 03103 L} = 0 , 0688 \frac{m o l}{L}$.
Since NaOH is monovalent its molarity is equal to its Normality =0,0688 (eq)/L.

For the oxalic acid (MM= 90,03 g/mol; ME = 45,01 g/eq) is necessary a double quantity of moles of NaOH, since oxalic acid is biprotic (1 eq = 2 mol), or (that is the same) the Equivalent Mass of the oxalic acid is the half of the Molecular Mass .
$e q {H}_{2} {C}_{2} {O}_{4} = \frac{0 , 1453 g}{45 , 01 \frac{g}{e q}} = 0 , 003228 e q$
$V \left(N a O H\right) = \frac{e q {H}_{2} {C}_{2} {O}_{4}}{N \left(N a O H\right)} = \frac{0 , 003228 e q}{0 , 0688 \frac{e q}{L}} = 0 , 0469 L = 46 , 9 m L$

Mar 11, 2017

$\left[N a O H \left(a q\right)\right] = 7.158 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We need (i) a stoichiometrically balanced equation for our primary standard:

${C}_{6} {H}_{4} \left(C {O}_{2}^{-} {K}^{+}\right) \left(C {O}_{2} H\right) + N a O H \rightarrow {C}_{6} {H}_{4} \left(C {O}_{2}^{-} {K}^{+}\right) \left(C {O}_{2}^{-} N {a}^{+}\right) + {H}_{2} O$

And this is a convienient 1:1 stoichiometry, so moles of phthalate is equivalent to moles of base:

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

i.e. $\text{Moles of KHP} = \frac{0.4536 \cdot g}{204.22 \cdot g \cdot m o {l}^{-} 1} = 2.221 \times {10}^{-} 3 m o l$.

Given the stoichiometry, this molar quantity was contained in the volume of titrant, i.e.

$\left[N a O H \left(a q\right)\right] = \text{Moles of base"/"Volume of titrant}$

$\frac{\frac{0.4536 \cdot g}{204.22 \cdot g \cdot m o {l}^{-} 1}}{\left(31.26 - 0.23\right) \times {10}^{-} 3 L} = 7.158 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$.

So we have $N a O H$, and we use to titrate a mass of oxalic acid. You must know this is a diacid, i.e. $H {O}_{2} C - C {O}_{2} H$, with a molecular mass of $90.03 \cdot g \cdot m o {l}^{-} 1$ (the dihydrate is much common, i.e. $H {O}_{2} C - C {O}_{2} H \cdot {\left(O {H}_{2}\right)}_{2}$, so if your results are off, question the quality of your oxalic acid, and you can assess this easily by melting point).

Again, we need a stoichiometric equation:

$\text{HO"_2"CCO"_2"H"(aq) + "2NaOH"(aq) rarr "Na"^(+)""^(-)"O"_2"CCO"_2^(-)"Na"^+ +2"H"_2"O} \left(l\right)$

$\text{Moles of oxalic acid}$ $=$ $\frac{0.1453 \cdot g}{90.03 \cdot g \cdot m o {l}^{-} 1} = 1.614 \times {10}^{-} 3 \cdot m o l$.

$\text{Volume of NaOH titrant required}$ $=$ (2xx1.614xx10^-3*mol)/(7.158xx10^-2*mol*L^-1)xx1000*mL*L^-1=??*mL.

And thus, we require under $50 \cdot m L$ of titrant, and this is a good volume for a titration.

This is a long problem, so I certainly would check the arithmetic. It is all too easy to make an error.