Question

How many milliliters of .75 M hydrochloric acid (HCl) are required to neutralize 60.0 ml of .3 M potassium hydroxide (KOH)?

Answer 1

Answer:

24 millilitres

Explanation:

The reaction between hydrochloric acid (HCl) and potassium hydroxide (KOH) is as following:

HCl + KOH => KCl + H2O

Here, Molarity of KOH = 0.3 M per liter

1L = 1000ml

60ml = 0.06 L

moles of KOH = 0.3 x 0.060  L = 0.018 mol (Normalize the volumes)

Molarity of HCL = 0.75 M per liter

Therefore, volume of HCl =  $\frac{0.018}{0.75}$ = 0.024 L

Hence, volume of HCl = 24 millilitres