Consider the titration of 80.0 mL Ba(OH)2 by 0.400 M HCl. Calculate the pH of the resulting solution after the following volumes of HCl have been added.
a.) 0.0 mL
*Which molecular equation do I use?
b.) 20.0mL
Ba(OH)2__+2HCl_+=>+_BaCl2_+____2HOH
80.0 mL__20.0mL
0.100M___0.400M
.008 mol_.008 mol
.004 left_ 0 left__.004 mol produced
pH = pKa
pH = -Log(1.05x10^-12)
pH = 11.98
*Did I do this one correctly?
2 answers
You didn't include the molarity of the Ba(OH)2.
0.100 M
7 answers