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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

80.0 mL__20.0mL
.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?

  • Chemistry -

    You didn't include the molarity of the Ba(OH)2.

  • Chemistry -

    0.100 M

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