The balanced chemical reaction for this is written as:
`2 HCl + Ba(OH)_2 -> BaCl_2 + H_2O `
`mol es OH- = (300)/(1000) L * 0.1 (mol es Ba(OH)_2)/(L) * (2 mol es OH^- )/(1 mol e Ba(OH)_2) `
`mol es OH- = 0.06 mol es `
`mol es H_3O^+ = (200)/(1000) L * 0.30 (mole s HCl)/(L) * (1 mol e H_3O^+)/(1 mol e HCl) `
`mol es H_3O^+ = 0.06 mol es `
`t otal volume = 300 + 200 = 500 mL = 0.5 L `
`mol es H_3O^+ = mol es OH^(-) - mol es H_3O^+ `
`mol es H_3O^+ = 0.06 - 0.06 = 0 mol e `
Concentration = 0/0.5L = 0 M
This means that it is a neutralization reaction where the there are no acid or base remaining in the reaction system.
Amount of `Ba(OH)_2` added `= (0.1xx300)/1000 = 0.03mol`
Amount of HCl added `= (0.3xx200)/1000 = 0.06mol`
`Ba(OH)_2+2HCl rarr BaCl_2+2H_2O`
Mole ratio according to reaction `Ba(OH)_2:HCl = 1:2`
Mole ratio of added components `Ba(OH)_2:HCl = 0.03:0.06 = 3:6 = 1:2`
So the reaction will take place completely without remaining any excess acid or base. The final solution will be a neutral one.
So the `H_3O^+` concentration will be same as the `H_3O^+` concentration of neutral water.
`2H_2O rarr H_3O^++OH^-`
Actually we cannot say anything about the `H_3O^+` concentration because the temperature is not given.
If the test temperature is 25C the concentration of `H_3O^+ = 10^(-7)M`
If the test temperature is less than 25C the concentration of `H_3O^+ < 10^(-7)M`
If the test temperature is higher than 25C the concentration of `H_3O^+ >10^(-7)M`
"Actually we cannot say anything about the `[H3O]^+` concentration" --- We can definitely say that. As there is no mention of the physical conditions of the experiment, it has to be the standard conditions, i.e. 1 atm. pressure and `25^oC` temperature.
What is more important, however, is the correct representation of dissociation process, which is a reversible phenomenon, and must be indicated accordingly.
Amount of base, in 300 ml of 0.1 M Ba(OH)2 solution=300*0.1*2=60 meq. (Ba(OH)2 is a diacidic base).
Amount of acid, in 200 ml of 0.3 M HCl solution=200*0.3=60 meq.
After neutralization reaction, there will be (60-60)=0 meq. of free HCl left. What remains after the reaction is only pure water. `H_3O^+` concentration of the resulting solution will be obtained from the dissociation of pure water (from the aqueous solution), such that, `[H_3O]^+=[OH]^-` in
`2H_2O stackrel larr rarr H_3O^++OH^-`
It therefore follows that,
`[H_3O]^+=sqrtK_w=sqrt(10^-14)=10^-7 M` (assuming the experimental temperature to be `25^oC` ).