Question

How many milliliters of 0.45 m hcl will neutralize 25.0 ml of 1.00 m koh?

Answer 1

Approx. 60⋅mL of hydrochloric acid are required.

Answer 2

Answer:

55.55 mL of HCl 0.45 m are needed to neutralize 25 mL of 1 m KOH

Explanation:

The equation that describes the reaction is:

$HCl + KOH = KCl + H_{2} O$

We can see that one mole of HCl is needed to consume one mole of KOH

Looking at the units:

$0.45 m HCl = \frac{0.45 MOL HCl}{Liter} \\1 m KOH = \frac{1 MOL KOH}{Liter}$

So, kepping in mind the previous statements, we proceed to calculate the total moles of KOH in the solution:

Now we know that we need 0.025 MOL of HCl to neutralize the solution; So:

$\frac{0.45 MOL HCl}{1 Liter} *\frac{1 Liter}{1000mLiters}=\frac{0.00045 MOL}{1mLiter}$

$\frac{0.025 MOL KOH}{\frac{0.00045 MOL HCL}{1mLiter} } = 55.5556 mLiters$