find the volume of 0.1 M hydrochloric acid required to react with 22.0ml of 0.07 M barium hydroxide
See the H2SO4/NaOH problem.
I really need help working them out they all look different to me
To find the volume of 0.1 M hydrochloric acid required to react with 22.0 ml of 0.07 M barium hydroxide, you need to use the concept of stoichiometry and the balanced chemical equation.
The balanced chemical equation for the reaction between hydrochloric acid (HCl) and barium hydroxide (Ba(OH)2) is:
2HCl + Ba(OH)2 -> BaCl2 + 2H2O
From the balanced equation, we can see that 2 moles of HCl react with 1 mole of Ba(OH)2 to form 1 mole of BaCl2 and 2 moles of water.
First, let's calculate the number of moles of barium hydroxide:
Moles of Ba(OH)2 = volume (in liters) x concentration (in moles per liter)
Moles of Ba(OH)2 = 22.0 ml x 0.07 mol/l / 1000 ml/l = 0.00154 moles
According to the balanced equation, 2 moles of HCl react with 1 mole of Ba(OH)2. Therefore, to completely react with 0.00154 moles of Ba(OH)2, you would need twice that amount of HCl.
Moles of HCl = 2 x 0.00154 = 0.00308 moles
Now, we can calculate the required volume of 0.1 M HCl:
Volume (in liters) = moles / concentration (in moles per liter)
Volume (in liters) = 0.00308 mol / 0.1 mol/l = 0.0308 liters
Finally, we need to convert the volume from liters to milliliters:
Volume (in milliliters) = 0.0308 liters x 1000 ml/l = 30.8 ml
Therefore, you would require approximately 30.8 ml of 0.1 M hydrochloric acid to react with 22.0 ml of 0.07 M barium hydroxide.