Question

Question #72f34

Answer 1

`[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`

Answer 2

`"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"`.