Respuesta :
Solution :
For the reaction :
[tex]$\text{TrisH}^+ + H_2O \rightarrow \text{Trish}^- + H_3O^+$[/tex]
we have
[tex]$Ka = \frac{[\text{Tris}^- \times H_3O]}{\text{Tris}^+}$[/tex]
  [tex]$=\frac{x^2}{0.02 -x}$[/tex]
 [tex]$=8.32 \times 10^{-9}$[/tex]
Clearing [tex]$x$[/tex], we have [tex]$x = 1.29 \times 10^{-5} \text{ moles of acid}$[/tex]
So to reach [tex]$\text{pH} = 7.8 (\text{pOH}= 14-7.8=6.2)$[/tex], one must have the [tex]$\text{OH}^-$[/tex] concentration of the :
[tex]$\text{[OH}^-]=10^{-pOH} = 6.31 \times 10^{-7} \text{ moles of base}$[/tex]
So we can add enough of 1 M NaOH in order to neutralize the acid that is calculated above and also adding the calculated base.
[tex]$\text {n NaOH}=1.29 \times 10^{-5}+6.31 \times 10^{-7}$[/tex]
      [tex]$= 1.35 \times 10^{-5} \text{ moles}$[/tex]
Volume NaOH [tex]$= 1.35 \times 10^{-5} \text{ moles} \times \frac{1000 \text{ mL}}{1 \text{ mol}} = 0.0135 \text{ mL}$[/tex]
Tris mass [tex]$H^+ = 0.02 \text{ mol} \times 157.6 \text{ g/mol}=3.152 \text{ g}$[/tex]
Now to prepare the said solution we must mix:
[tex]$3.152 \text{ g Tris H} + 0.0135 \text{ mL NaOH} \ 1 M$[/tex] gauge to 1000 mL with water.
