Question

What is the molarity of HCl found in a titration where 30 ml of HCl is titrated with 15 ml of .1M NaOH?

Answer 1

`NaOH(aq) + HCl(aq) -> NaCl(aq) + H_2O(l)`

The Molarity of HCl is `.05M`

Explanation:

Since in the balanced equation, NaOH and HCl have a `1:1` ratio they must have the same number of mole

We can say that mols is equal to `(Molarity)/(Liters)` since the units cancel out to equal moles `((mol)/(L))/(L)` or `(mol * L)/L` or `mol`
Finally, we can set up a ratio to solve for the molarity of HCl

`(.1_"M")/(15_"ml") = (x_"M")/(30_"ml")`

This comes out to `x=.05_"M"` or `HCl=.05_"M"`

Answer 2

I get `"0.05 M"`, which is a reasonable answer.

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Write the chemical reaction. This neutralization is of a strong base and strong acid, and is a double-replacement reaction:

`"NaOH"(aq) + "HCl"(aq) -> "H"_2"O"(l) + "NaCl"(aq)`

This is `1:1`, i.e. `"1 mol OH"^(-)` reacted with `"1 mol H"^(+)`. Therefore, what we are looking for is for equal mols of `"HCl"` to react with `"NaOH"`.

`"0.1 mol NaOH"/cancel"L" xx 15 "m"cancel"L" = "1.5 mmols NaOH"`

`= "1.5 mmols OH"^(-)`

This reacts exactly with `"1.5 mmols H"^(+)`, or `"1.5 mmols HCl"`.

If this is contained in `"30 mL"` of aqueous `"HCl"` solution, then the molar concentration is:

`color(blue)(["HCl"]) = (1.5 cancel"m""mols HCl")/(30 cancel"m""L solution") = color(blue)("0.05 M HCl")`