Answer :
To solve this problem, we need to find out the volume of 0.894 M HCl required to completely react with 36.2 mL of 0.831 M NaOH.
The chemical reaction equation given is:
[tex]\[ \text{HCl} (aq) + \text{NaOH} (aq) \rightarrow \text{NaCl} (aq) + \text{H}_2\text{O} (l) \][/tex]
This is a simple acid-base neutralization reaction where one mole of HCl reacts with one mole of NaOH. This means the molar ratio between HCl and NaOH is 1:1.
We will use the formula for titrations:
[tex]\[ M_1 \times V_1 = M_2 \times V_2 \][/tex]
Where:
- [tex]\( M_1 \)[/tex] is the molarity of HCl (0.894 M),
- [tex]\( V_1 \)[/tex] is the volume of HCl we want to find,
- [tex]\( M_2 \)[/tex] is the molarity of NaOH (0.831 M),
- [tex]\( V_2 \)[/tex] is the volume of NaOH (36.2 mL).
Since the reaction ratio is 1:1, we can apply the formula directly:
[tex]\[
0.894 \times V_1 = 0.831 \times 36.2
\][/tex]
Now, solve for [tex]\( V_1 \)[/tex]:
[tex]\[
V_1 = \frac{0.831 \times 36.2}{0.894}
\][/tex]
This calculation will give us the volume [tex]\( V_1 \)[/tex]:
[tex]\[
V_1 \approx 33.6489932885906
\][/tex]
Rounded to one decimal place, this is approximately 33.6 mL.
So, the volume of 0.894 M HCl required to titrate 36.2 mL of 0.831 M NaOH is [tex]\( e. \)[/tex] 33.6 mL.
The chemical reaction equation given is:
[tex]\[ \text{HCl} (aq) + \text{NaOH} (aq) \rightarrow \text{NaCl} (aq) + \text{H}_2\text{O} (l) \][/tex]
This is a simple acid-base neutralization reaction where one mole of HCl reacts with one mole of NaOH. This means the molar ratio between HCl and NaOH is 1:1.
We will use the formula for titrations:
[tex]\[ M_1 \times V_1 = M_2 \times V_2 \][/tex]
Where:
- [tex]\( M_1 \)[/tex] is the molarity of HCl (0.894 M),
- [tex]\( V_1 \)[/tex] is the volume of HCl we want to find,
- [tex]\( M_2 \)[/tex] is the molarity of NaOH (0.831 M),
- [tex]\( V_2 \)[/tex] is the volume of NaOH (36.2 mL).
Since the reaction ratio is 1:1, we can apply the formula directly:
[tex]\[
0.894 \times V_1 = 0.831 \times 36.2
\][/tex]
Now, solve for [tex]\( V_1 \)[/tex]:
[tex]\[
V_1 = \frac{0.831 \times 36.2}{0.894}
\][/tex]
This calculation will give us the volume [tex]\( V_1 \)[/tex]:
[tex]\[
V_1 \approx 33.6489932885906
\][/tex]
Rounded to one decimal place, this is approximately 33.6 mL.
So, the volume of 0.894 M HCl required to titrate 36.2 mL of 0.831 M NaOH is [tex]\( e. \)[/tex] 33.6 mL.
