Respuesta :
Answer:
For a: The volume of HCl needed is 4.52 mL
For b: The volume of HCl needed is 25.63 mL
For c: The volume of HCl needed is 43.33 mL
Explanation:
To calculate the volume of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base
- For a:
We are given:
[tex]n_1=1\\M_1=0.105M\\V_1=?mL\\n_2=1\\M_2=9.50\times 10^{-2}M=0.00950M\\V_2=50.0mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.105\times V_1=1\times 0.00950\times 50\\\\V_1=\frac{1\times 0.00950\times 50}{1\times 0.105}=4.52mL[/tex]
Hence, the volume of HCl needed is 4.52 mL
- For b:
We are given:
[tex]n_1=1\\M_1=0.105M\\V_1=?mL\\n_2=1\\M_2=0.117M\\V_2=23.0mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.105\times V_1=1\times 0.117\times 23\\\\V_1=\frac{1\times 0.117\times 23}{1\times 0.105}=25.63mL[/tex]
Hence, the volume of HCl needed is 25.63 mL
- For c:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
Given mass of NaOH = 1.40 g
Molar mass of NaOH = 40 g/mol
Volume of solution = 1 L
Putting values in above equation, we get:
[tex]\text{Molarity of NaOH solution}=\frac{1.40}{40\times 1}\\\\\text{Molarity of NaOH}=0.035M[/tex]
We are given:
[tex]n_1=1\\M_1=0.105M\\V_1=?mL\\n_2=1\\M_2=0.035M\\V_2=130mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.105\times V_1=1\times 0.035\times 130\\\\V_1=\frac{1\times 0.035\times 130}{1\times 0.105}=43.33mL[/tex]
Hence, the volume of HCl needed is 43.33 mL
a: The volume of HCl is 4.52 mL
b: The volume of HCl is 25.63 mL
c: The volume of HCl is 43.33 mL
Calculation for volume of acid:
It is given by the formula for dilution or neutralization reaction:
n₁*M₁*V₁= n₂*M₂*V₂
where,
n₁, M₁ ,V₁ are the n-factor, molarity and volume of acid
n₂M₂ ,V₂ are the n-factor, molarity and volume of base
For a:
Given:
n₁=1
M₁=0.105M
n₂ =1
M₂=0.00950M
V₂= 50.0mL
To find: V₁=?
On substituting the values:
n₁*M₁*V₁= n₂*M₂*V₂
1*0.105*V₁=1*0.00905*50
V₁=4.52mL
Hence, the volume of HCl needed is 4.52 mL
For b:
Given:
n₁=1
M₁=0.105M
n₂ =1
M₂=0.117 M
V₂= 23.0 mL
To find: V₁=?
On substituting the values:
n₁*M₁*V₁= n₂*M₂*V₂
1*0.105*V₁=1*0.117 * 23
V₁=25.63 mL
Hence, the volume of HCl needed is 25.63 mL
For c:
Given:
Mass of NaOH = 1.40 g
Molar mass of NaOH = 40 g/mol
Volume of solution = 1 L
On substituting the values:
Molarity of NaOH = 1.40 / 40
Molarity of NaOH = 0.035M
Given:
n₁=1
M₁=0.105M
n₂ =1
M₂=0.035 M
V₂= 130 mL
To find: V₁=?
On substituting the values:
n₁*M₁*V₁= n₂*M₂*V₂
1*0.105*V₁=1*0.035 * 130
V₁=43.33 mL
Hence, the volume of HCl needed is 43.33 mL.
Find more information about Molarity here:
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