Question

250 ml of 0.5 m NaOH is added to 500 mL of 1 M HCl. What is the final concentration of the remaining HCl?

Answer 1

Neutralization Reaction:

1. Reaction in which acid and base are mixed together to form a salt.
2. Example: $\mathrm{KOH}\hspace{0.17em}+\mathrm{HBr}\to \mathrm{KBr}+{\mathrm{H}}_2\mathrm{O}$

Molarity:

1. Molarity is defined as the number of moles of solute present in 1L of solution.
2. Formula : $M=\frac{n}{V}$

where; M = molarity of the solution

n = number of moles of solute

V = volume of solution

Explanation:

1. Reaction of NaOH with HCl: $\mathrm{NaOH}+\hspace{0.17em}\mathrm{HCl}\to \mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}$
2. It is given that 250ml of 0.5M NaOH used was used. Thus, $0.5=\frac{n_{NaOH}}{\left({\displaystyle \frac{250}{1000}}\right)}$ ${\mathrm{n}}_{\mathrm{NaOH}}$ = 0.125 moles
3. It is given that 500ml of 1M HCl was used. Thus, $1=\frac{n_{HCl}}{\left({\displaystyle \frac{500}{1000}}\right)}$ ${\mathrm{n}}_{\mathrm{HCl}}$= 0.5 moles
4. According to reaction 1 mole of NaOH requires 1 mole of HCl.
5. So, 0.125 moles of NaOH requires 0.125 moles of HCl.
6. Thus, number of moles of HCl remaining after reaction are: n = ${\mathrm{n}}_{\mathrm{HCl}}$ - ${\mathrm{n}}_{\mathrm{NaOH}}$ = 0.5-0.125 = 0.375 moles
7. Molarity of final solution is: number of moles of HCl left = 0.375 moles volume of final solution = 250ml + 500ml = 750ml Thus, $M=\frac{0.375}{\left({\displaystyle \frac{750}{1000}}\right)}=0.5$
8. Thus the final concentration of the remaining HCl = 0.5M