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

When pH = 9.50 ; Volume = 14.8 L

When pH = 4.70 ; Volume = 4.51 L

Step-by-step 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= ([H_2A^-][H_3O^+])/([H_3A])`

`1.0*10^(-3)= ([H_2A^-][H_3O^+])/([H_3A])`

`(1.0*10^(-3))/([H_3O^+])= ([H_2A^-])/([H_3A])`

where;
`[H_3O^+] = 10^(-4.70)`

so, we have:

`(1.0*10^(-3))/([10^(-4.70)])= ([H_2A^-])/([H_3A])`

`([H_2A^-])/([H_3A])= 10^(1.7)`

`([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^-)+ (n_(H_2A^-))/(50.12) = 1.3328*10^(-2)`

`n_(H_2A^-)= (1.3328*10^(-2))/((1+(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 =
`(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
`(7.30+11.70)/(2)`

=
`(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 =
`(2*1.3328*10^(-2))/(1.8)`

= 0.1481 L

= 14.8 L

When pH = 9.50 ; Volume = 14.8 L