Question

Write the balanced equation. Then calculate the volume of 0.65 M HCl required to completely neutralize 400.0 ml of 0.88 M KOH.

Answer 1

Hello!

The balanced equation for the neutralization of KOH is the following:

HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l)

To calculate the volume of HCl required, we can apply the following equation:

 $moles HCl = moles KOH \\ \\ cHCl*vHCl=cKOH*vKOH \\ \\ vHCl= \frac{cKOH*vKOH}{cHCl}= \frac{400 mL*0,88M}{0,65M}= 541,54mL$

So, the required volume of HCl is 541,54 mL

Have a nice day!

Answer 2

Answer:

HCl + KOH → KCl + H2O

.54 L or 540 mL are required

Explanation:

0.400 l × 0.88 M KOH = 0.35 moles KOH

.35 * 1 mole of HCl/1mole of KOH= .35 moles of HCl

.35/.65= .54L

Edmentum