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:
[tex]HCl + KOH = KCl + H_{2} O[/tex]
We can see that one mole of HCl is needed to consume one mole of KOH
Looking at the units:
[tex]0.45 m HCl = \frac{0.45 MOL HCl}{Liter} \\1 m KOH = \frac{1 MOL KOH}{Liter}[/tex]
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:
[tex]\frac{0.45 MOL HCl}{1 Liter} *\frac{1 Liter}{1000mLiters}=\frac{0.00045 MOL}{1mLiter}[/tex]
[tex]\frac{0.025 MOL KOH}{\frac{0.00045 MOL HCL}{1mLiter} } = 55.5556 mLiters[/tex]
