What is the #pH# of the final solution...?

A #25*mL# volume of #HBr(aq)# at #0.050*mol*L^-1# concentration is mixed with a #10*mL# volume of #KOH(aq)# at #0.020*mol*L^-1# concentration. What is the #pH# of the final solution?

1 Answer
Mar 23, 2017

Answer:

A stoichiometric equation is required:

#HBr(aq) + KOH(aq) rarr NaBr(aq) + H_2O(l)#

We get (finally) #pH=1.52#

Explanation:

And thus we need to find the amount of substance of both #"hydrogen bromide"#, #"hydrobromic acid"#, and #"potassium hydroxide"#.

We use the relationship, #"Concentration"="Moles of solute"/"Volume of solution"#, OR

#"Concentration"xx"Volume"="Moles of solute"#

#"Moles of HBr"=25xx10^-3cancelLxx0.050*mol*cancel(L^-1)=1.25xx10^-3*mol.#

#"Moles of KOH"=10xx10^-3cancelLxx0.020*mol*cancel(L^-1)=0.200xx10^-3*mol.#

Note that I converted the #mL# volume to #L# by using the relationship: #1*mL-=1xx10^-3*L#

Clearly, the hydrobromic acid is in excess. And given 1:1 stoichiometry, there are #((1.25*mol-0.200*mol)xx10^-3)/(35xx10^-3L)# #HBr# remaining.

So #[HBr]=((1.25*mol-0.200*mol)xx10^-3)/(35xx10^-3L)#

#=(1.05*molxx10^-3)/(35xx10^-3L)=0.030*mol*L^-1# with respect to #HBr#.

#pH=-log_10[H_3O^+]=-log_10(0.030)=1.52#