What volume (L) of 0.256 M potassium hydroxide solution would just neutralize 61.3 ml of 0.167 M H2SO4 solution?
To find the volume of potassium hydroxide solution needed to neutralize a given volume of sulfuric acid solution, you can use the concept of stoichiometry.
In this case, we need to determine the volume of the potassium hydroxide solution, so we can start by writing a balanced chemical equation for the reaction between potassium hydroxide (KOH) and sulfuric acid (H2SO4):
2 KOH + H2SO4 → K2SO4 + 2 H2O
From the balanced equation, we can see that 2 moles of KOH react with 1 mole of H2SO4. Therefore, the molar ratio between KOH and H2SO4 is 2:1.
Now, let's calculate the moles of H2SO4 in the given volume of 0.167 M H2SO4 solution. We can use the formula:
Moles = Concentration (M) × Volume (L)
Moles of H2SO4 = 0.167 M × 0.0613 L = 0.01023 moles
Since the molar ratio between KOH and H2SO4 is 2:1, we need twice the number of moles of KOH to neutralize the H2SO4. Therefore, the moles of KOH required would be:
Moles of KOH = 2 × Moles of H2SO4 = 2 × 0.01023 moles = 0.02046 moles
Finally, we can calculate the volume of the 0.256 M KOH solution using the formula:
Volume (L) = Moles / Concentration (M)
Volume of KOH = 0.02046 moles / 0.256 M = 0.07992 L = 79.92 mL
Therefore, the volume of the 0.256 M potassium hydroxide solution needed to neutralize 61.3 mL of 0.167 M H2SO4 solution is approximately 79.92 mL.