How many milliliters of a 0.100 M NaOH solution are required to neutralize 25.0 milliliters of 0.150 M HCl?
Select one:
a. 25.0 ml
b. 37.5 ml
c. 125 ml
d. 167 ml
mols HCl = M x L = ?
mols NaOH = mols HCl
M NaOH = mols NaOH/L NaOH. You know mols and M, solve for L and convert to mL.
m(1) * v(1) = m(2) * v(2)
125ml
37.5 ml
the formula for all of these types of questions in Chemistry is: M(1)*V(1) = M(2)*V(2)
So...
0.100M*V(1) = 0.150M*25.0ml
V(1) = (0.150M*25.0ml)/0.100M
V(1) = 37.5
You can use M(1)*V(1) = M(2)*V(2) for all your questions in neutralization. It will always work.
To find the number of milliliters of the NaOH solution required to neutralize the HCl, we need to use the concept of stoichiometry.
First, let's write the balanced chemical equation for the neutralization reaction between NaOH and HCl:
NaOH + HCl → NaCl + H2O
From this equation, we can see that 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of NaCl and 1 mole of water.
Now, let's use the given concentrations of the solutions to find the moles of HCl present in the 25.0 mL:
Molarity (M) = moles/volume (L)
Given:
Concentration of HCl = 0.150 M
Volume of HCl = 25.0 mL = 25.0/1000 = 0.025 L
Moles of HCl = 0.150 M × 0.025 L = 0.00375 moles
Since the stoichiometry of the balanced equation is 1:1, we can conclude that the number of moles of NaOH required to neutralize the HCl is also 0.00375 moles.
Now let's determine the volume of the 0.100 M NaOH solution needed to provide 0.00375 moles of NaOH:
Molarity (M) = moles/volume (L)
Given:
Concentration of NaOH = 0.100 M
Moles of NaOH = 0.00375 moles
Volume of NaOH = Moles/Concentration = 0.00375 moles/0.100 M = 0.0375 L = 37.5 mL
Therefore, the answer is b. 37.5 ml.