Answer :

200 mL of 2.00 M HCl should be added to 100 mL of 2.00 M KOH to produce a solution with a pH of 7.

To determine the amount of 2.00 M HCl needed to produce a solution with a pH of 7 when mixed with 100 mL of 2.00 M KOH. we need to consider the stoichiometry of the acid-base neutralization reactions

The balanced equation for the reaction between HCl and KOH is:

HCl + KOH → H2O + KCl

In this neutralization reaction, 1 mole of HCl reacts with 1 mole of KOH to produce 1 mole of water and 1 mole of KCl.

Since we want the resulting solution to have a pH of 7, it should be neutral. Given that the initial volume of KOH is 100 mL and its concentration is 2.00 M, we can calculate the moles of KOH:

Moles of KOH = Volume (in liters) × Concentration

Moles of KOH = 0.100 L × 2.00 mol/L

To achieve neutrality, we need an equal number of moles of HCl. Therefore, the amount of HCl needed is also 0.100 L × 2.00 mol/L.

Converting the volume to milliliters:

Volume of HCl needed = 0.100 L × 2.00 mol/L = 200 mL

Learn more about acid-base neutralization reaction here:

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