Question

How do you calculate the pH of the solution made by adding 0.50 mol of `HOBr` and 0.30 mol of `KOBr` to 1.00 L of water?

The value of Ka for `HOBr` is `2.0*10^{-9}`.

Answer 1

You can do it like this:

Explanation:

`HOBr` dissociates:

`HOBr_((aq))rightleftharpoonsH_((aq))^(+)+OBr_((aq))^-`

The expression for `K_a` is:

`K_a=([H_((aq))^+][OBr_((aq))^-])/([HOBr_((aq))])`

These are equilibrium concentrations.

To find the `pH` we need to know the `H_((aq))^+` concentration so rearranging gives:

`[H_((aq))^(+)]=K_axx[[HOBr_((aq))]]/[[OBr_((aq))^-]]`

Because the value of `K_a` is so small we can see that the position of equilibrium lies well to the left.

This means that the initial moles given will be a very close approximation to the equilibrium moles so we can use them in the expression.

There will be a volume change on adding these substances to water so the final volume will not now be 1 litre.

This does not matter as the volume is common to both so cancels out:

`:.[H_((aq))^+]=2xx10^(-9)xx0.5/cancel(V)/0.3/cancel(V)=3.33xx10^(-9)" ""mol/l"`

`pH=-log[H_((aq))^+]`

`:.pH=-log[3.33xx10^(-9)]`

`color(red)(pH=8.47)`