Final answer:
To find the volume of 0.196 M Ba(OH)2 needed to react with 53.9 mL of 0.144 M HI, first calculate the moles of HI, then determine the moles of Ba(OH)2 required, and finally calculate the volume of Ba(OH)2 solution. After performing the calculations, 19.8 mL of Ba(OH)2 is needed.
Step-by-step explanation:
To determine the volume of 0.196 M Ba(OH)2 solution needed to completely react with 53.9 mL of a 0.144 M HI solution, first we need to write the balanced chemical equation:
Ba(OH)2(aq) + 2HI(aq) → BaI2(aq) + 2H2O(l)
From the equation, we see that one mole of Ba(OH)2 reacts with two moles of HI. Next, we calculate the moles of HI:
Moles of HI = volume (in L) × molarity (M) = 0.0539 L × 0.144 M = 0.0077616 mol
Since we need half the amount of moles of Ba(OH)2 to react with HI, we then have:
Moles of Ba(OH)2 required = 0.0077616 mol / 2 = 0.0038808 mol
Finally, we use the molarity of Ba(OH)2 to find the volume required:
Volume of Ba(OH)2 = moles / molarity = 0.0038808 mol / 0.196 M = 0.0198 L
Converting liters to milliliters:
Volume of Ba(OH)2 = 0.0198 L × 1000 mL/L = 19.8 mL
Therefore, 19.8 mL of the 0.196 M Ba(OH)2 solution is needed to completely react with 53.9 mL of 0.144 M HI solution.