How much 0.150 M sodium hydroxide would be required to just neutralize 15 ml of 0.175 M hydrochloric acid?
moles HCl = M x L = ??
moles HCl = moles NaOH because the reaction is 1 mole to 1 mole. Write the equation.
Then M NaOH = moles/L.
Solve for L (and convert to mL if desired.)
To find out how much 0.150 M sodium hydroxide would be required to neutralize 15 mL of 0.175 M hydrochloric acid, we can use the concept of stoichiometry.
Stoichiometry involves using the balanced chemical equation and the molar ratios to calculate the amount of reactants needed to fully react.
First, let's write the balanced chemical equation for the reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl):
NaOH + HCl -> NaCl + H2O
From the balanced equation, we can see that the molar ratio between NaOH and HCl is 1:1, meaning that 1 mole of NaOH reacts with 1 mole of HCl.
Now, let's calculate the number of moles of HCl present in the 15 mL of 0.175 M HCl solution:
Molarity (M) = moles/volume (L)
0.175 M = moles/0.015 L
moles of HCl = 0.175 M * 0.015 L
moles of HCl = 0.002625 mol
Since the molar ratio between NaOH and HCl is 1:1, we need an equal number of moles of NaOH to neutralize the HCl.
Therefore, the amount of NaOH required can be calculated by multiplying the number of moles of HCl by the molarity (0.150 M) of NaOH:
moles of NaOH = 0.150 M * 0.002625 mol
moles of NaOH = 0.00039375 mol
Finally, to convert moles to volume, we can use the equation:
Volume (L) = moles/Molarity
Volume of NaOH = 0.00039375 mol / 0.150 M
Volume of NaOH = 0.002625 L or 2.625 mL
Therefore, approximately 2.625 mL of 0.150 M sodium hydroxide would be required to neutralize 15 mL of 0.175 M hydrochloric acid.