How do you calculate the pH of the solution made by adding 0.50 mol of $H O B r$ $HOBr$ and 0.30 mol of $K O B r$ $KOBr$ to 1.00 L of water?

Question

The value of Ka for $H O B r$ $HOBr$ is $2.0 \cdot 10^{- 9}$ $2.0*10^{-9}$ .

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Answer 1 · 2016-06-26T14:22:22

You can do it like this:

$H O B r$ $HOBr$ dissociates:

$H O B r_{(a q)} ⇌ H_{(a q)}^{+} + O B r_{(a q)}^{-}$ $HOBr_((aq))rightleftharpoonsH_((aq))^(+)+OBr_((aq))^-$

The expression for $K_{a}$ $K_a$ is:

$K_{a} = \frac{[H_{(a q)}^{+}] [O B r_{(a q)}^{-}]}{[H O B r_{(a q)}]}$ $K_a=([H_((aq))^+][OBr_((aq))^-])/([HOBr_((aq))])$

These are equilibrium concentrations.

To find the $p H$ $pH$ we need to know the $H_{(a q)}^{+}$ $H_((aq))^+$ concentration so rearranging gives:

$[H_{(a q)}^{+}] = K_{a} \times \frac{[H O B r_{(a q)}]}{[O B r_{(a q)}^{-}]}$ $[H_((aq))^(+)]=K_axx[[HOBr_((aq))]]/[[OBr_((aq))^-]]$

Because the value of $K_{a}$ $K_a$ 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:

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

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

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

$color(red)(pH=8.47)$