The molarity of the 500 mL KOH solution needed to neutralize the 40 mL of 2.50 M HCl is 0.2 M
Balanced equation
HCl + KOH —> KCl + H₂O
From the balanced equation above,
The mole ratio of the acid, HCl (nA) = 1
The mole ratio of the base, KOH (nB) = 1
How to determine the molarity of KOH
- Volume of acid, HCl (Va) = 40 mL
- Molarity of acid, HCl (Ma) = 2.5 M
- Volume of base, KOH (Vb) = 500 mL
- Molarity of base, KOH (Mb) =?
MaVa / MbVb = nA / nB
(2.5 × 40) / (0.15 × Vb) = 1
100 / (Mb × 500) = 1
Cross multiply
Mb × 500 = 100
Divide both side by 500
Mb = 100 / 500
Mb = 0.2 M
Thus, the molarity of the KOH solution needed is 0.2 M
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