A 35.0 mL sample of 0.225 M #HBr# was titrated with 42.3 mL of #KOH#. What is the concentration of the #KOH#?

1 Answer
anor277
Apr 26, 2016

Approx. #0.19*mol*L^-1#

Explanation:

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

Clearly there is a #1:1# equivalence between acid and base.

#"Moles of HBr"=35.0xx10^-3Lxx0.225*mol*L^-1=7.88xx10^-3*mol.#

#[KOH]# #=# #"Moles of base"/"volume of solution"#

#=# #(7.88xx10^-3*mol)/(42.3xx10^-3*L)# #=# #??*mol*L^-1#

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