A 8.71 g sample of an aqueous solution of hydroiodic acid contains an unknown amount of the acid.
If 25.0 mL of 0.388 M barium hydroxide are required to neutralize the hydroiodic acid, what is the percent by weight of hydroiodic acid in the mixture?
I understand how to solve this. The only problem I am having is coming up with the balanced equation. Please help, thank you!!
To determine the balanced equation for the reaction between hydroiodic acid (HI) and barium hydroxide (Ba(OH)2), you need to first identify the reactants and products involved.
The balanced equation can be obtained by combining the ionic equations for the dissociation of both hydroiodic acid and barium hydroxide in water.
The dissociation of hydroiodic acid (HI) can be represented by the equation:
HI(aq) → H+(aq) + I-(aq)
The dissociation of barium hydroxide (Ba(OH)2) can be represented by the equation:
Ba(OH)2(aq) → Ba2+(aq) + 2OH-(aq)
Now, considering the neutralization reaction between hydroiodic acid and barium hydroxide, the balanced equation can be written as:
2HI(aq) + Ba(OH)2(aq) → BaI2(aq) + 2H2O(l)
In this reaction, two moles of hydroiodic acid react with one mole of barium hydroxide to form one mole of barium iodide and two moles of water.
Now that you have the balanced equation, you can proceed to solve the problem.
Remember, acid + base = salt + water
Ba(OH)2 + 2HI ==> 2H2O + BaI2.