Question

What volume of 0.378 m hydroiodic acid (hi) is required to neutralize 21.6 ml of 0.520 m barium hydroxide, ba(oh)2?

Answer 1

The volume of o.378 M of Hl is calculated as follows
find the moles of Ba(OH)2 used

that is moles = molarity x volume /1000
= 21.6 x 0.520/1000= 0.0112 moles

write the equation for reaction

Ba(oh)2 +2Hl ------> Bal2 + 2H2O

fro the equation above the reacting ratio between Ba(OH)2 to Hl is 1:2 therefore the moles of Hl = 2 x0.0112= 0.0224 moles of Hl

volume of HCl = moles of Hl / molarity of Hl x1000

that is 0.0224/0.378x1000= 59.3 ml

Answer 2

first, we have to get moles of Ba(OH)2 that reacts with HCl according to this equation:
Ba(OH)2(aq) + 2HCl (aq) → BaCl2(aq) + 2H2O(l)

when moles of Ba(OH)2 = molarity * volume
= 0.52 M * 0.0216 L
= 0.011232 moles
from the balanced equation we can see that the mole ratio between
HCl : Ba(OH)2 = 2:1
So moles of HCl = 2* moles of Ba(OH)2
= 2* 0.011232 moles
= 0.0225 moles
now, we can get the volume:
when volume = no of moles of HCl / molarity of HCl
= 0.0225 moles / 0.378 M
∴volume = 0.0595 L = 59.5 mL