Calculate the volume of NaOH 0.5 mol/L necessary to neutralize 300 mL of HCl 0.2 M.
na= how do I calculate this?
Ma= 0.2 M
Va= 0.300 L
nb= 0.5 mol
Mb= 0.5 M
Vb= ?
Ma x Va = Mb x Vb
Substitute and solve for the unknown.
.12
The neutralization reaction of NaOH and HCl is given by
H
C
l
(
a
q
)
+
N
a
O
H
(
a
q
)
→
N
a
C
l
(
a
q
)
+
H
2
O
(
l
)
HCl (aq)+NaOH (aq)→NaCl (aq)+H2O (l)
We can employ a ratio-and-proportion method in calculating the volume of NaOH:
0.5
M
N
a
O
H
x
m
L
N
a
O
H
=
0.2
M
H
C
l
300
m
L
H
C
l
x
=
750
m
L
N
a
O
H
0.5 M NaOHx mL NaOH=0.2 M HCl300 mL HClx=750 mL NaOH
∴
V
N
a
O
H
=
750
m
L
∴ VNaOH=750 mL
To calculate the volume of NaOH 0.5 mol/L necessary to neutralize 300 mL of HCl 0.2 M, you can use the concept of stoichiometry in a chemical reaction.
First, let's write the balanced chemical equation for the reaction between NaOH and HCl:
NaOH + HCl → NaCl + H2O
From the equation, we can see that one mole of NaOH reacts with one mole of HCl to produce one mole of NaCl and one mole of water.
Now, let's use the concept of stoichiometry to find the volume of NaOH solution required. We can start by equating the number of moles of HCl and NaOH:
Moles of HCl = (Molarity of HCl) × (Volume of HCl in liters)
Moles of NaOH = Moles of HCl
Given:
Molarity of HCl (Ma) = 0.2 M
Volume of HCl (Va) = 0.300 L
Moles of NaOH (nb) = 0.5 mol
Using the equation:
Moles of HCl = Ma × Va
Substituting the given values:
Moles of HCl = 0.2 M × 0.300 L
Calculate the moles of HCl:
Moles of HCl = 0.06 mol
Since one mole of NaOH reacts with one mole of HCl, we can say that the moles of NaOH required to neutralize the HCl is also 0.06 mol.
Next, we can use the molarity and molality relationship to calculate the volume of NaOH solution required:
Molarity of NaOH (Mb) = 0.5 M
Moles of NaOH (nb) = 0.06 mol
Volume of NaOH solution (Vb) = ?
Using the equation:
Molarity = Moles / Volume
Rearranging the equation:
Volume = Moles / Molarity
Substituting the given values:
Volume of NaOH solution (Vb) = 0.06 mol / 0.5 M
Calculate the volume of NaOH solution required:
Vb = 0.12 L or 120 mL
So, the volume of NaOH 0.5 mol/L necessary to neutralize 300 mL of HCl 0.2 M is 120 mL.