30.0 mL of 0.2 M of Ba(OH)2 is required to neutralize 25 ml of H3C6H5O7 . What is the molarity of the H3C6H5O7 ?
a. 0.167 M
b. 0.55 M
c. 0.25 M
d. 0.75 M
e. 0.084 M
not sure if this is the right answer, but here's how i'd do it:
you have 30.0 mL of .2 M Ba(OH)2, which is .030 L of .2 M, or .006 mol of Ba(OH)2. But since there's 2 moles of OH- in that, that's .012 mol of OH- needed to neutralize the H+ in the acid.
I'm assuming from the structure of the acid (with the 3 H in the front) that there are 3 H+ per molecule of acid, or 3 mole H+ per mole of acid. So if there's .012 mole of OH- to neutralize the H+, the acid has 1/3 that many moles.
.012/3 = .004 moles of acid / .025 L = .16 M, so I'd say A
hope this isn't embarrassingly super wrong!