Respuesta :
The pH of a buffer =7.38.
Steps
This equation can be written first:
[tex]H_{3} PO_{4} (aq)+ OH- > H_{2} PO_{4} (aq)+H_{2}O(l)[/tex]
We can now determine how many moles of phosphoric acid there are:
n(H3PO4)= 0.5L×1M=0.5mol= n(H2PO4−)
There is 0.3mol of NaOH left after the 1:1 reaction between H3PO4 and NaOH (0.8mol - 0.5mol = 0.3mol).
H2PO4 and the remaining 0.3 mol of NaOH react.
Since H2PO4 and NaOH react in a 1:1 ratio, 0.2 mol of NaOH is left over (0.5 mol - 0.3 mol = 0.2 mol).
Ka (H2PO4) = [tex]6.3$\times10{^{-8}}$ pKa[/tex]= -log Ka
=7.20
The Henderson-Hasselbalch equation can now be used:
[tex]PH= PKa +log\frac{A}{HA}[/tex]
PH=7.20+[tex]\frac{0.3mol/ 0.5L}{0.2mol/ 0.5L}[/tex]
PH = 7.38
Describe pH buffer.
The equilibrium constant Ka of the weak acid and the ratio of weak base (A-) to weak acid (HA) in solution are the two variables that determine the pH of a buffer.
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