I have seen in a table that pKa of H3PO4 =2.12 so pH =(pKa-log c)/2 =(2.12-
log0.2)/2=1.41 at equilibrium pH =pKa=2.12
Solution
I have seen in a table that pKa of H3PO4 =2.12 so pH =(pKa-log c)/2 =(2.12-
log0.2)/2=1.41 at equilibrium pH =pKa=2.12.
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I have seen in a table that pKa of H3PO4 =2.12 s.pdf
1. I have seen in a table that pKa of H3PO4 =2.12 so pH =(pKa-log c)/2 =(2.12-
log0.2)/2=1.41 at equilibrium pH =pKa=2.12
Solution
I have seen in a table that pKa of H3PO4 =2.12 so pH =(pKa-log c)/2 =(2.12-
log0.2)/2=1.41 at equilibrium pH =pKa=2.12