Question

The titration of 0.02500 L of a diprotic acid solution with 0.1000 M NaOH requires 34.72 mL of titrant to reach the second equivalence point. The pH is 3.95 at the first equivalence point and 9.27 at the second equivalence point. If the add solution contained 0.2015 g of the acid, what is the molar mass, pK_a1, and pK_a2 of the acid?

Answer 1

Answer:

molar mass (of the diprotic acid) = 116g/mol

$pK_{a_1}= 1.83$

$pK_{a_2}=6.07$

Explanation:

The number of moles of $NaOH$ and $H^+$ are equal at the equivalence point since they are both taking part in the diprotic acid.

0.1 M means 0.1 moles in 1L or 0.1 moles in 1000 mL

Number of moles of 0.1 M NaOH at final equivalence point ;

=  $34.72 mL * \frac{0.1mole}{1000mL}$

= 0.00347 moles of NaOH

However, the number of moles of the diprotic acid in the 0.25 L solution is = $\frac {0.00347}{2 }$      (due to the fact that half of the concentration of NaOH is needed to give the same amount of  $H^+$

$= 0.001736$  

Given that:

The  Total acid in the solution = 0.2015 g; to calculate the molar mass ; we have :

no of moles = $\frac{mass}{molar mass}$

molar mass= $\frac{mass}{numbers of moles}$

molar mass = $\frac{0.2015}{0.001736}$

molar mass (of the diprotic acid) = 116g/mol

At the second equivalence point;

The pH = 9.27

pH = $-log[H^+]$

$[H^+]= 10^{-pH}$

$[H^+] = 10^{-9.27$

$[H+] = 5.37 * 10^{-10$

$[OH^-]$ can be calculated as follows:

$\frac{10^{-14}}{[H^+]} = \frac {10^{-14}}{5.37* 10^{-10}}$

$[OH^-]$ = $1.86*10^{-5}$

Let represent the equation from the reaction after the second equivalence point with:

$X^{2-} + H_2O \rightleftharpoons HX^- + OH^-$

where:

$X^{2-}$ =  $\frac{0.001736}{(25 + 34.72)} * 1000$

$X^{2-}$ =  $0.029 M$

 The ICE Table is shown as follows;

                      $X^{2-}$    $+$    $H_2O$       $\rightleftharpoons$      $HX^-$     $+$        $OH^-$

Initial            0.029                                  0                     0

Change        -x                                         +x                   +x  

Equilibrium  (0.029 - x)                            x                     x

$Kb = \frac{[HX^-][OH^-]}{[X^{2-}]}$

$Kb = \frac{x^2}{0.029-x}$

$k_b}= \frac{(1.86*10^{-5})^2}{0.029-(1.86*10^{-5})}$

$x = 1.191*10^{-8}$

$K_a} = \frac {10^{-14}}{K_b}$

$K_a} = \frac {10^{-14}}{1.191*10^{-8}}$

$K_a} = 8.39 * 10^{-7}$

$pK_a} = -log K_a$

$pK_a} = -log (8.39*10^{-7})$

$pK_{a_2}=6.07$

At the first equivalence point we have all H2X getting converted to  $HX^-$  $HX^-$ is an amphoteric species which implies that it can serve as both an acid and a base

As such, in this process:

$pH = pK_{a_1} + \frac{pKa_2}{2}$

Given that: the pH = 3.95

Then;

$3.95 = pK_{a_1} + \frac{6.07}{2}$

$3.95*2 = pK_{a_1} +{6.07}$

$7.9 = pK_{a_1} +{6.07}$

$pK_{a_1}= 7.9 - {6.07}$

$pK_{a_1}= 1.83$

Answer 2

Answer:

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Explanation: