Question

Consider a 0.586 M aqueous solution of barium hydroxide- Ba(OH)2 (aq)

1. How many grams Ba(OH)2 are dissolved in 0.191 dL of 0.586 M Ba(OH)2 (aq)
2. How many individual hydroxide ions (OH-1) are found in 13.4 mL of 0.586 M Ba(OH)2 (aq)
3. What volume in L of 0.586 M Ba(OH)2 (aq) contains 0.466 OUNCES of Ba(OH)2 dissolved in it?
4. If 16.0 mL of water are added to the 31.5 mL of 0.586 M Ba(OH)2 what is the new solutions molarity?
5. Suppose you had titrated your vinegar sample with barium hydroxide instead of sodium hydroxide:

BA(OH)2 (aq) a 2HC2H3O2 (aq)--> Ba(C2H3O2)2 (aq) + 2H2O(l)
What volume in mL of 0.586 M Ba(OH)2 (aq) must be added to a 5.00 mL sample of vinegar to reach the equivalence point.  You can just use X to represent the Molarity of vinegar.

 

Answer 1