how much 0.150 M sodium hydroxide would be required to just neutralize 15 ml of 0.175 M hydrochloric acid?
moles HCl = grams/molar mass
moles HCl = moles NaOH
M NaOH = moles/L.
To determine the amount of sodium hydroxide required to neutralize the hydrochloric acid, we can use the concept of stoichiometry and the balanced chemical equation between sodium hydroxide (NaOH) and hydrochloric acid (HCl):
NaOH + HCl -> NaCl + H2O
From the balanced equation, we can see that 1 mole of sodium hydroxide reacts with 1 mole of hydrochloric acid, forming 1 mole of sodium chloride and 1 mole of water.
First, we need to calculate the number of moles of hydrochloric acid present in the given solution. To do this, we use the formula:
moles = concentration (M) x volume (L)
Given: Concentration of hydrochloric acid (HCl) = 0.175 M
Volume of hydrochloric acid (HCl) = 15 mL (convert to liters by dividing by 1000)
moles of HCl = 0.175 M x (15 mL / 1000 mL/L)
moles of HCl = 0.002625 moles
Since the stoichiometric ratio between HCl and NaOH is 1:1, we know that 0.002625 moles of sodium hydroxide will be required to neutralize the hydrochloric acid.
Finally, we need to calculate the volume of 0.150 M sodium hydroxide solution required to contain 0.002625 moles of sodium hydroxide using the equation:
volume (L) = moles / concentration (M)
Given: Concentration of sodium hydroxide (NaOH) = 0.150 M
Moles of sodium hydroxide (NaOH) = 0.002625 moles
volume (NaOH) = 0.002625 moles / 0.150 M
volume (NaOH) = 0.0175 L (convert to milliliters by multiplying by 1000)
volume (NaOH) = 17.5 mL
Therefore, approximately 17.5 mL of 0.150 M sodium hydroxide solution would be required to just neutralize 15 mL of 0.175 M hydrochloric acid.