Thread: chemistry

Calculate the hydroxide ion concentration in a 0.25 mol/L HBr(aq) solution.

Originally Posted by euclid2

Calculate the hydroxide ion concentration in a 0.25 mol/L HBr(aq) solution.

HBr(aq) ---> H+(aq)+Br-(aq)

entities are H+(aq),Br-(aq),H2O(l)

Kw=[H+(aq)][OH-(aq)]

[OH-(aq)]=Kw/[H+(aq)]

now simply make the necessary substitutions

Originally Posted by TheMasterMind

HBr(aq) ---> H+(aq)+Br-(aq)

entities are H+(aq),Br-(aq),H2O(l)

Kw=[H+(aq)][OH-(aq)]

[OH-(aq)]=Kw/[H+(aq)]

now simply make the necessary substitutions

I'm not sure where you're getting this from can you expand please?

Originally Posted by euclid2

I'm not sure where you're getting this from can you expand please?

[OH-(aq)]=Kw/[H+(aq)]

=1.00*10^-14/0.25

=4.0*10^-14 mol/L

[OH-(aq)]=4.0*10^-14mol/L

Does this make sense?

Originally Posted by TheMasterMind

[OH-(aq)]=Kw/[H+(aq)]

=1.00*10^-14/0.25

=4.0*10^-14 mol/L

[OH-(aq)]=4.0*10^-14mol/L

Does this make sense?

It seems you are ignoring the amount of [OH-(aq)]

Originally Posted by euclid2

It seems you are ignoring the amount of [OH-(aq)]

The amount of OH is insignificant produced by the ionization of water in presence of Br

Originally Posted by euclid2

I'm not sure where you're getting this from can you expand please?

Note: since Kw=[H+(aq)][OH-(aq]

then [H+(aq)]=Kw/[OH-(aq)]

and [OH-(aq)]=Kw/[H+(aq)]