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Sagot :
Solution :
The reaction :
[tex]$\text{TrisH}^+ + \text{H}_2\text{O} \rightarrow \text{Tris}^- +\text{H}_3\text{O}^+$[/tex]
We have
[tex]$K_a = \frac{[\text{Tris}^-]\times[\text{H}_3\text{O}^-]}{[\text{TrisH}^+]}$[/tex]
[tex]$=\frac{x^2}{0.02-x}$[/tex]
[tex]$= 8.32 \times 10^{-9}$[/tex]
Clearing x, we have [tex]$x=1.29 \times 10^{-5}$[/tex] moles of acid
Now to reach pH = [tex]$7.8 (\text{ pOH} = 14-7.8 = 6.2)$[/tex], we must have an [tex]$OH^-$[/tex] concentration of
[tex]$[OH^-] = 10^{-pOH}$[/tex]
[tex]$=10^{-6.2}$[/tex]
[tex]$=6.31 \times 10^{-7}$[/tex] moles of base
We must add enough NaOH of 1 M to neutralize the acid calculated above and also add the calculated base.
[tex]$n \ NaOH = 1.29 \times 10^{-5} + 6.31 \times 10^{-7}$[/tex]
[tex]$=1.35 \times 10^{-5}$[/tex] moles
Vol [tex]$NaOH = 1.35 \times 10^{-5} \text{ moles} \times \frac{1000 \ mL}{1 \ mol}$[/tex]
= 0.0135 L
Tris mass [tex]$H^+ = 0.02 \text{ mol} \times 157.6 \ g/mol$[/tex]
= 3.152 g
To prepare the said solution we must mix
-- [tex]$3.152 \ g \text{ TrisH}^+$[/tex]
-- [tex]$0.0135 \ mL \ NaOH \ 1M$[/tex]
-- [tex]$\text{Gauge to 1000 mL with H}_2\text{O}$[/tex]
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