Calculate amount of milliliters of 2.0 M HCl needed to neutralize 0.15 mol of NaOH
Start with a balanced equation
HCl + NaOH -> NaCl
so one mole of HCl reacts with 1 mole of NaOH
Thus we need to find the volume of HCl that contains 0.15 mole of HCl
We can do this my simple proportion
1 litre (or 1000 ml) of HCl contains 2.0 moles
1 ml of HCl containes 0.002 moles
0.5 ml contains 0.001 moles
0.5 ml x 150 contains 0.150 moles
To find the amount of milliliters of 2.0 M HCl needed to neutralize 0.15 mol of NaOH, we can use the balanced chemical equation and stoichiometry.
The balanced chemical equation for the neutralization reaction between HCl and NaOH is:
HCl + NaOH -> NaCl + H2O
From the equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH. Therefore, the mole ratio between HCl and NaOH is 1:1.
To find the amount of HCl needed, we can use the equation:
moles of solute = molarity × volume (in liters)
We can rearrange this equation to solve for volume:
volume (in liters) = moles of solute / molarity
First, let's convert the given amount of NaOH from moles to liters:
moles of NaOH = 0.15 mol
Next, let's determine the volume of 2.0 M HCl needed:
volume of HCl (in liters) = moles of NaOH / molarity of HCl
volume of HCl (in liters) = 0.15 mol / 2.0 moles/L
Finally, let's convert the volume from liters to milliliters:
volume of HCl (in milliliters) = volume of HCl (in liters) × 1000
volume of HCl (in milliliters) = (0.15 mol / 2.0 moles/L) × 1000
volume of HCl (in milliliters) = 75 milliliters
Therefore, 75 milliliters of 2.0 M HCl is needed to neutralize 0.15 moles of NaOH.