What is the pH of a solution obtained by mixing 30.00 mL of 0.500 M HCl and 30.00 mL of 0.250 M NaOH?
#mol HCl = 0.50M * 0.03L = 0.015 mol
#mol NaOH = 0.25M * 0.03L = 0.0075mol
HCl clearly has the more moles.
Thus, the excess number of moles of
0.015-0.0075 = 0.0075 mol
#M HCl = 0.0075mol/0.06L = 0.125M HCl
(Since 0.06L used since total volume
of both solutions is 30mL )
We then know that
pH = -log[H3O+] = -log[HCl]
pH = -log(0.125)= 2.079441542=2.08
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