Question
What is the [OH-] in the final solution prepared by mixing 20.0 mL of 0.50 M HCl with 30.0mL of 0.10 Ba(OH)2?
-
0.10 M
-
0.40 M
-
0.0050 M
-
0.12 M
Solution
A.
0.10 M
Number of milliequivalents of HCl = 2 x 0.050 x 1 = 1
Number of milliequivalents of Ba(OH)2 = 2 x 30 x 0.10 = 6
[OH]- of final soluton
milliequivalent of Ba(OH)2
