Respuesta :
Question is incomplete, complete question is;
A 34.8 mL solution of [tex]H_2SO_4[/tex] (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).
[tex]H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)[/tex]
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of [tex]H_2SO_4(aq)[/tex]? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.
Answer:
0.044 M is the molarity of [tex]H_2SO_4[/tex](aq).
Explanation:
The reaction taking place here is in between acid and base which means that it is a neutralization reaction .
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_4[/tex]
[tex]n_1,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=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL[/tex]
Putting values in above equation, we get:
[tex]2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M[/tex]
0.044 M is the molarity of [tex]H_2SO_4[/tex](aq).
