43.96 mL.
Hydrochloric acid HCl reacts with sodium chloride NaOH at a 1:1 ratio.
[tex]\text{HCl} +\text{NaOH} \to \text{NaCl}+\text{H}_2\text{O}[/tex].
[tex]V(\text{HCl)} = \dfrac{n(\text{HCl})}{c(\text{HCl})} \\\phantom{V(\text{HCl)}} = \dfrac{n(\text{NaOH})}{c(\text{HCl})} \\\phantom{V(\text{HCl)}} = \dfrac{c(\text{NaOH}) \cdot V(\text{NaOH})}{c(\text{HCl})} \\\phantom{V(\text{HCl)}} = V(\text{NaOH}) \cdot \dfrac{c(\text{NaOH})}{c(\text{HCl})}\\\phantom{V(\text{HCl)}} = 25.00\times \dfrac{0.800}{0.455} \\\phantom{V(\text{HCl)}} = 43.96\;\text{mL}[/tex].
Answer: The volume of HCl solution required is 44.0 mL
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 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=0.455M\\V_1=?mL\\n_2=1\\M_2=0.800M\\V_2=25.00mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.455\times V_1=1\times 0.800\times 25.00\\\\V_1=\frac{1\times 0.800\times 25.00}{1\times 0.455}=44.0mL[/tex]
Hence, the volume of HCl solution required is 44.0 mL
