Answer:
When pH = 9.50 ; Volume = 14.8 L
When pH = 4.70 ; Volume = 4.51 L
Explanation:
We all know that:
[tex]pKa = -logKa[/tex]
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 [tex]pKa_1[/tex] ; we have [tex]-logKa_1[/tex]
= [tex]-log(1.0*10^{-3})[/tex]
= 3.00
[tex]pKa_2 = -logKa_2[/tex]
= [tex]-log(5.0*10^{-3})[/tex]
= 7.30
[tex]pKa_3 = -logKa_3[/tex]
= [tex]-log(2.0*10^{-12})[/tex]
= 11.70
At pH = 4.70:
The pH (4.70) is closer to [tex]pKa_1[/tex] than [tex]pKa_2[/tex] and [tex]pKa_3[/tex].
However, the only important pKa for the dissociation of the acid will be directed towards only [tex]pKa_1[/tex]
So: At pH = 4.70
-log [tex][H_3O^+][/tex] = pH = 4.70
[tex][H_3O^+] = 10^{-4.70}[/tex]
[tex]H_3A_{(aq)}[/tex] [tex]+[/tex] [tex]H_2O_{(l)}[/tex] ⇄ [tex]H_2A^-}_{(aq)}[/tex] [tex]+[/tex] [tex]H_3O^+_{(aq)}[/tex]
[tex]Ka_1= \frac{[H_2A^-][H_3O^+]}{[H_3A]}[/tex]
[tex]1.0*10^{-3}= \frac{[H_2A^-][H_3O^+]}{[H_3A]}[/tex]
[tex]\frac{1.0*10^{-3}}{[H_3O^+]}= \frac{[H_2A^-]}{[H_3A]}[/tex]
where; [tex][H_3O^+] = 10^{-4.70}[/tex]
so, we have:
[tex]\frac{1.0*10^{-3}}{[10^{-4.70}]}= \frac{[H_2A^-]}{[H_3A]}[/tex]
[tex]\frac{[H_2A^-]}{[H_3A]}= 10^{1.7}[/tex]
[tex]\frac{[H_2A^-]}{[H_3A]}= 50.12[/tex]
Given that;
283.0 mL of [tex]5.60*10^{-2}M[/tex] solution is given:
Then ; 283.0 mL = 0.283 L
However;
[tex]n_{(H_2A^-)} + n_{(H_3A)[/tex] = [tex]0.283[/tex] [tex]*5.60*10^{-2}M[/tex]
= 0.013328
= [tex]1.3328*10^{-2} M[/tex]
[tex]n_{H_2A^-}+ \frac{n_{H_2A^-}}{50.12} = 1.3328*10^{-2}[/tex]
[tex]n_{H_2A^-}= \frac{1.3328*10^{-2}}{(1+\frac{1}{50.12} )}[/tex]
= 0.01307 mole
moles of NaOH required to convert [tex]H_3A[/tex] to [tex]H_2A^-[/tex] i.e [tex](n_{H_2A^-})[/tex]
= [tex]1.307*10^{-2}moles[/tex]
Finally; the volume of NaOH required = [tex]\frac{1.307*10^{-2}}{2.90}[/tex]
= 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 [tex]pKa_2[/tex] and [tex]pKa_3[/tex] yields 9.50
i.e [tex]\frac{7.30+11.70}{2}[/tex]
= [tex]\frac{19}{2}[/tex]
= 9.50
Here; virtually all, [tex]H_3A[/tex] is in [tex]H_2A^-[/tex] form;
SO; the moles of NaOH required to convert [tex]H_3A[/tex] to [tex]H_2A^-[/tex] will be:
= 2 × initial [tex]H_3A[/tex] = volume of NaOH × 1.8 M
= 2 × 1.3328 × 10⁻² mol = Volume of NaOH × 1.8 M
Volume of NaOH = [tex]\frac{2*1.3328*10^{-2}}{1.8}[/tex]
= 0.1481 L
= 14.8 L
When pH = 9.50 ; Volume = 14.8 L
