Question

In the titration of 238.0 mL of a 5.60×10-2 M solution of acid H3A (Ka1 = 1.0×10-3, Ka2 = 5.0×10-8, Ka3 = 2.0×10-12), calculate the volume of 2.90 M NaOH required to reach the following pH values. pH = 9.50 Volume required = mL pH = 4.70 Volume required = mL

Answer 1

Answer:

When pH = 9.50 ; Volume = 14.8 L

When pH = 4.70 ; Volume = 4.51 L

Explanation:

We all know that:

$pKa = -logKa$

So let us first calculate the pKa values of the following Ka:

Given that:

Ka1 = 1.0×10⁻³

Ka2 = 5.0×10⁻⁸

Ka3 = 2.0×10⁻¹²

For $pKa_1$ ; we have $-logKa_1$

= $-log(1.0*10^{-3})$

= 3.00

$pKa_2 = -logKa_2$

= $-log(5.0*10^{-3})$

= 7.30

$pKa_3 = -logKa_3$

= $-log(2.0*10^{-12})$

= 11.70

At pH = 4.70:

The pH (4.70) is closer to $pKa_1$ than $pKa_2$ and $pKa_3$.

However, the only important pKa for the dissociation of the acid will be directed towards only  $pKa_1$

So: At pH = 4.70

-log $[H_3O^+]$ = pH = 4.70

$[H_3O^+] = 10^{-4.70}$

$H_3A_{(aq)}$      $+$     $H_2O_{(l)}$     ⇄     $H_2A^-}_{(aq)}$     $+$     $H_3O^+_{(aq)}$

$Ka_1= \frac{[H_2A^-][H_3O^+]}{[H_3A]}$

$1.0*10^{-3}= \frac{[H_2A^-][H_3O^+]}{[H_3A]}$

$\frac{1.0*10^{-3}}{[H_3O^+]}= \frac{[H_2A^-]}{[H_3A]}$

where; $[H_3O^+] = 10^{-4.70}$

so, we have:

$\frac{1.0*10^{-3}}{[10^{-4.70}]}= \frac{[H_2A^-]}{[H_3A]}$

$\frac{[H_2A^-]}{[H_3A]}= 10^{1.7}$

$\frac{[H_2A^-]}{[H_3A]}= 50.12$

Given that;

283.0 mL of $5.60*10^{-2}M$ solution is given:

Then ; 283.0 mL = 0.283 L

However;

$n_{(H_2A^-)} + n_{(H_3A)$ = $0.283$ $*5.60*10^{-2}M$

= 0.013328

= $1.3328*10^{-2} M$

$n_{H_2A^-}+ \frac{n_{H_2A^-}}{50.12} = 1.3328*10^{-2}$

$n_{H_2A^-}= \frac{1.3328*10^{-2}}{(1+\frac{1}{50.12} )}$

= 0.01307 mole

moles of NaOH required to convert $H_3A$ to $H_2A^-$  i.e $(n_{H_2A^-})$

= $1.307*10^{-2}moles$

Finally; the volume of NaOH required = $\frac{1.307*10^{-2}}{2.90}$

= 0.004507

= 4.51 mL

When pH = 4.70 ; Volume = 4.51 L

When pH = 9.50

We try to understand that the average of $pKa_2$ and $pKa_3$ yields 9.50

i.e $\frac{7.30+11.70}{2}$

=  $\frac{19}{2}$

= 9.50

Here; virtually all, $H_3A$  is in $H_2A^-$  form;

SO; the moles of NaOH required to convert $H_3A$ to $H_2A^-$ will be:

= 2 × initial  $H_3A$ = volume of NaOH × 1.8 M

= 2 × 1.3328 × 10⁻² mol = Volume  of NaOH × 1.8 M

Volume  of NaOH  = $\frac{2*1.3328*10^{-2}}{1.8}$

= 0.1481 L

= 14.8 L

When pH = 9.50 ; Volume = 14.8 L