Respuesta :
Answer: Molarity of [tex]H_2SO_3[/tex] is 1.2 M.
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 which is [tex]H_2SO_3[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Mg(OH)_2[/tex]
We are given:
[tex]n_1=2\\M_1=?\\V_1=250mL\\n_2=2\\M_2=1.5M\\V_2=200mL[/tex]
Putting values in above equation, we get:
[tex]2\times M_1\times 250=2\times 1.5\times 200\\\\M_1=1.2M[/tex]
Thus molarity of [tex]H_2SO_3[/tex] is 1.2 M if 250 mL of the acid was used to neutralize 200 mL of 1.5 M solution of [tex]Mg(OH)_2[/tex]
