What volume of 0.900 M HCl is required to complete neutralize 50.0 ml of a 0.500 M Ba(OH)2 solution?
balance the equation. You will see that it takes two moles of HCl for each mole of Ba(OH)2
2molesHCl=1moleBa(OH)2
2*.9*Va=1*.5*50ml
solve for volume of acid, Va
To calculate the volume of HCl required to completely neutralize the Ba(OH)2 solution, we need to use the balanced chemical equation for the reaction between HCl and Ba(OH)2:
2HCl + Ba(OH)2 -> BaCl2 + 2H2O
From the equation, we can see that 2 moles of HCl react with 1 mole of Ba(OH)2 to produce 1 mole of BaCl2 and 2 moles of water.
First, let's calculate the number of moles of Ba(OH)2 in the 50.0 ml of 0.500 M solution:
moles of Ba(OH)2 = volume (L) × concentration (M)
= 0.050 L × 0.500 M
= 0.025 moles
Since the molar ratio is 2:1 between HCl and Ba(OH)2, we can see that we need twice as many moles of HCl to completely neutralize the Ba(OH)2. Therefore, we need 0.050 moles of HCl.
Now, let's determine the required volume of 0.900 M HCl solution to deliver 0.050 moles of HCl:
volume (L) = moles of HCl / concentration (M)
= 0.050 moles / 0.900 M
= 0.056 L
Finally, we need to convert the volume from liters to milliliters:
volume (ml) = volume (L) × 1000
= 0.056 L × 1000
= 56 ml
Therefore, you would need 56 ml of the 0.900 M HCl solution to completely neutralize the 50.0 ml of 0.500 M Ba(OH)2 solution.