how many ml of a.15M NaOH solution are required to neutralize 35.00 ml of 0.22M Hcl
mLHCl x MHCl = mLNaOH x MNaOH
51.33 mL
To determine how many milliliters (ml) of a 0.15M NaOH solution are required to neutralize 35.00 ml of 0.22M HCl, we can use the concept of stoichiometry.
First, write the balanced chemical equation for the reaction between NaOH and HCl:
NaOH + HCl -> NaCl + H2O
From the equation, we can see that the stoichiometric ratio between NaOH and HCl is 1:1. That means one mole of NaOH reacts with one mole of HCl.
To find the number of moles of HCl in 35.00 ml of 0.22M HCl, we can use the formula:
moles = volume (in liters) x concentration (in moles per liter)
moles of HCl = 0.035 L x 0.22 mol/L
moles of HCl = 0.0077 mol
Since the stoichiometric ratio is 1:1, we need the same number of moles of NaOH to neutralize the HCl.
Now, we can determine the volume of the 0.15M NaOH solution required to neutralize the HCl.
volume (in liters) = moles / concentration (in moles per liter)
volume (in liters) = 0.0077 mol / 0.15 mol/L
volume (in liters) = 0.0513 L
Finally, convert the volume to milliliters:
volume (in ml) = 0.0513 L x 1000 ml/L
volume (in ml) = 51.3 ml
Therefore, you would need approximately 51.3 ml of a 0.15M NaOH solution to neutralize 35.00 ml of 0.22M HCl.