Question #72f34

2 Answers
Dec 10, 2017

#[HCl]~=0.065*mol*L^-1#

Explanation:

We first write the stoichiometric equation....

#HCl(aq) + KOH(aq) rarr KCl(aq) +H_2O(l)#

...to establish the 1:1 equivalence...

And thus #[HCl]=(25.81*mLxx10^-3*L*mL^-1xx0.1250*mol*L^-1)/(50.00xx10^-3*L)#

#=0.06453*mol*L^-1#

Dec 10, 2017

#"0.065 M"#

Explanation:

The formula for the reaction is

#"HCl" + "KOH" -> "KCl" + "H"_2"O"#

and its already balanced (yea!).

When the mL and M of something #("KOH")# is givenm use cross multiplication to find how many mols (remember #"M"# is #"mol/liter"#, or #"mol/1000 mL"#):

#"0.1250 mol"/"1000 mL" = "x mol"/ "25.81 mL"#

to get #3.23xx10^-3# #"moles KOH"#, and since in the reaction equation all coefficients are 1, that means there were #3.23xx10^-3# #"moles"# of #"HCl"# in the #"50 mL"# sample.

Then, you use cross multiplication again to find how many moles would be in #"1000 mL"# of that sample:

#"x mol"/"1000 mL" = (3.23xx10^-3 "mol")/ "50 mL"#

which will give you #"0.065 mol/1000 mL"#, which is #"0.065 M"#.