Answer: The molar concentration of sulfuric acid in the original sample is 1.943 M
Explanation:
To calculate the molarity 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_4[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=2\\M_1=?\\V_1=56.5mL\\n_2=1\\M_2=0.5824M\\V_2=43.37mL[/tex]
Putting values in above equation, we get:
[tex]2\times M_1\times 56.5=1\times 0.5824\times 43.37[/tex]
[tex]M_1=0.2235[/tex]
Now to calculate the molarity of original solution:
[tex]M_1\times 6.5=0.2235\times 56.5[/tex]
[tex]M_1=1.943[/tex]
Thus the molar concentration of sulfuric acid in the original sample is 1.943 M
