Answer:
0.20 M is the approximate molarity of the acid used
Explanation:
Concentration of base = [tex]C_1=0.45 M[/tex]
Volume of the base used in titration = [tex]V_1=10 mL[/tex]
Concentration of an acid = [tex]C_2=?[/tex]
Volume of an acid used in titration = [tex]V_2=22 mL[/tex]
Molar ratio = [tex]\frac{m_1}{m_2}=\frac{1}{1}[/tex]
[tex]m_1=m_2=1[/tex]
[tex]m_1C_1V_1=m_2C_2V_2[/tex] ( neutralization )
[tex]C_2=\frac{m_1C_1V_1}{m_2V_2}=\frac{1\times 0.45 M\times 10 mL}{1\times 22 mL}=0.20 M[/tex]
0.20 M is the approximate molarity of the acid used
