What is the H3O+ concentration of a solution that is formed by combining 300ml of 0.1 M Ba(OH)2 with 200 ml of 0.3 M Hcl.

Expert Answers
jeew-m eNotes educator| Certified Educator

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`

jerichorayel eNotes educator| Certified Educator

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. 

llltkl | Student

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` ).

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