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}#.

1 Answer
Jun 26, 2016

Answer:

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)#