In: Chemistry
200 mL* 0.10 M H2PO4- = 200*0.1 = 20
millimols.
----mL* 1.0 M NaOH = ---- millimoles.
H2PO4- + OH- ==>
HPO4- + H2O
20
0
0
initial
-x
-x
x
change
20-x
0
x
equilibrium
pH = pKa + log[(base)/(acid)]
7.50 = 6.86 + log(x/20-x)
x = 16.27 mL
7.50 = 6.86 + log(x/20-x)