Respuesta :
Answer:
0.0445 M is the initial concentration of sulfurous acid.
Explanation:
[tex]H_2SO_3+2NaOH\rightarrow Na_2SO_3+2H_2O[/tex]
To calculate the concentration 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 which is [tex]H_2SO_3[/tex]
are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=2\\M_1=?\\V_1=25.00mL\\n_2=1\\M_2=0.13 M\\V_2=17.12 mL[/tex]
Putting values in above equation, we get:
[tex]2\times M_1\times 25.00 mL=1\times 0.13 M\times 17.12 mL[/tex]
[tex]x=\frac{1\times 0.13 M\times 17.12 mL}{2\times 25.00}=0.0445 M[/tex]
0.0445 M is the initial concentration of sulfurous acid.
Steps included :
0.0445 M is the preliminary attention of sulfurous acid.
H 2 SO 3 +2NaOH Na 2 SO 3 +2H 2 O
To calculate the attention of acid, we use the equation given through neutralization reaction:
[tex]n1M1V1 = n2M2V2[/tex]
where,
[tex]n1M1V1[/tex]are the n-factor, molarity, and extent of acid that is [tex]H2SO3[/tex]
are the n-factor, molarity, and extent of the base that is [tex]NaOH[/tex]
We are given:
[tex]n1 = 2[/tex]
[tex]M1=?[/tex]
[tex]V1 = 25.00 mL[/tex]
[tex]n2= 1\\M2= 0.13 M\\v2 = 17.12 mL[/tex]
Putting values in the above equation, we get:
[tex]2M_ * 25.00 * m * L = 1 * 0.1 3M > 17.12mL[/tex]
[tex]0.31M[/tex]
[tex]0.0445 M[/tex] is the initial concentration of sulfurous acid.
What is concentration?
Concentrated sulfuric acid is a weak acid (see Acids and Bases) and has relatively few ions at room temperature, resulting in inadequate electrolytes. When it is cold, it does not easily react with common metals such as iron and copper.
Hence it concluded that the above equation is the initial concentration of sulfurous acid
To know more about the initial concentration of sulfurous acid refer to the link :
https://brainly.com/question/26178014
