Question

what volume of a 0.455 M HCl solution is required to neutralize 25.00 mL of a 0.800 M sodium hydroxide solution

Answer 1

Answer: The volume of HCl solution required is 44.0 mL

Step-by-step explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

`n_1M_1V_1=n_2M_2V_2`

where,

`n_1,M_1\text{ and }V_1` are the n-factor, molarity and volume of acid which is HCl

`n_2,M_2\text{ and }V_2` are the n-factor, molarity and volume of base which is NaOH.

We are given:

`n_1=1\\M_1=0.455M\\V_1=?mL\\n_2=1\\M_2=0.800M\\V_2=25.00mL`

Putting values in above equation, we get:

`1* 0.455* V_1=1* 0.800* 25.00\\\\V_1=(1* 0.800* 25.00)/(1* 0.455)=44.0mL`

Hence, the volume of HCl solution required is 44.0 mL

Answer 2

43.96 mL.

Explanation

Hydrochloric acid HCl reacts with sodium chloride NaOH at a 1:1 ratio.

`\text{HCl} +\text{NaOH} \to \text{NaCl}+\text{H}_2\text{O}`.

`V(\text{HCl)} = \frac{n(\text{HCl})}{c(\text{HCl})} \\\phantom{V(\text{HCl)}} = \frac{n(\text{NaOH})}{c(\text{HCl})} \\\phantom{V(\text{HCl)}} = \frac{c(\text{NaOH}) \cdot V(\text{NaOH})}{c(\text{HCl})} \\\phantom{V(\text{HCl)}} = V(\text{NaOH}) \cdot \frac{c(\text{NaOH})}{c(\text{HCl})}\\\phantom{V(\text{HCl)}} = 25.00* (0.800)/(0.455) \\\phantom{V(\text{HCl)}} = 43.96\;\text{mL}`.