To determine the volume of hydrochloric acid required to react with the caustic potash (potassium hydroxide), we need to use the balanced chemical equation for the reaction between potassium hydroxide (KOH) and hydrochloric acid (HCl):
KOH + HCl → KCl + H₂O
From the balanced equation, we can see that the stoichiometric ratio between KOH and HCl is 1:1. This means that 1 mole of KOH reacts with 1 mole of HCl.
To calculate the volume of hydrochloric acid required, we need to know the concentration of the hydrochloric acid solution. You mentioned it is a 1 molar (1 M) solution, which means it contains 1 mole of HCl per liter of solution.
Let's assume we have 0.4 grams of caustic potash (KOH). To determine the amount of KOH in moles, we need to divide the mass by the molar mass of KOH, which is approximately 56.11 g/mol.
Number of moles of KOH = 0.4 g / 56.11 g/mol
Now, since the stoichiometric ratio is 1:1, the number of moles of HCl required will be the same as the number of moles of KOH.
Number of moles of HCl required = Number of moles of KOH = 0.4 g / 56.11 g/mol
Since we have a 1 M HCl solution, which means it contains 1 mole of HCl per liter of solution, the volume of HCl required can be calculated using the following equation:
Volume of HCl required (in liters) = Number of moles of HCl required
Therefore, the volume of hydrochloric acid required to complete the reaction is equal to the number of moles of HCl required, which is 0.4 g / 56.11 g/mol or 0.0071 moles. This also translates to 0.0071 liters or 7.1 milliliters of hydrochloric acid solution.