Question

Calculate the volume of $1.00mol{L}^{-1}$ aqueous sodium hydroxide that is neutralized by $200mL$ of $2.00mol{L}^{-1}$ aqueous hydrochloric acid and the mass of sodium chloride produced. Neutralization reaction is:

$NaOH(aq.)+HCl(aq.)\to NaCl(aq.)+H_{2}O\left(l\right).$

Answer 1

Meq. of $NaOH=$ Meq. of $HCl=$Meq. of $NaCl$

$N\times V_{mL}=N\times V_{mL}(N_{HCl}=M_{HCl};N_{NaOH}=M_{NaOH})$

$1\times V=2\times 200$

$\therefore V=400mL$

Meq. of $NaCl=2\times 200=400$

$\therefore \frac{w}{58.5}\times 1000=400$

$\therefore W_{NaCl}=23.4g$