The question is incomplete, here is the complete question.
The end point in a titration of a 50.00-mL sample of aqueous HCl was reached by addition of 35.23 mL of 0.250 M NaOH titrant. The titration reaction is:
[tex]HCl(aq.)+NaOH(aq.)\rightarrow NaCl(aq.)+H_2O(l)[/tex]
What is the molarity of the HCl?
Answer: The molarity of HCl solution is 0.176 M
Explanation:
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 HCl
[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=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.250M\\V_2=35.23mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 50.00=1\times 0.250\times 35.23\\\\M_1=\frac{1\times 0.250\times 35.23}{1\times 50.00}=0.176M[/tex]
Hence, the molarity of HCl solution is 0.176 M
The molarity of the acid would be 0.178 M
The equation of the reaction would be as follows:
HCl + NaOH ---> NaCl + H2O
Volume of acid (Va) used = 50 mL
Volume of base (Vb) = 35.23
Molarity of base (Cb) = 0.250 M
Molarity of acid (Ca)= ?
Using the equation: CaVa/CbVb = Na/Nb = 1/1
Ca = 0.250 x 35.23/50
= 0.178 M
More on acid/base titration calculations can be found here: https://brainly.com/question/6389903?referrer=searchResults
