Answer:
A: NaOH + HCl = NaCl +H2O
B: Volume of NaOH = 22.2ml
C: [tex]mole=3.33\times10^{-3}[/tex]
D: Volume of HCl = 9.45 ml
E: [tex]molarity=0.35 mole/litre = final concentration of HCl[/tex]
Explanation:
Part A: Balanced chemical equaion
NaOH + HCl = NaCl +H2O
part B:
Volume of NaOH used in titration is 22.2 ml because volume is taken upto the end point of the titration.
Part C:
Calculation of moles of NaOH used:
[tex]mole=volume in litre\times molarity[/tex]
[tex]mole=\frac{22.2}{1000} \times0.15[/tex]
[tex]mole=3.33\times10^{-3}[/tex]
Part D:
Volume of HCl used in titration is 9.45 ml because volume is taken upto the end point of the titration.
Part E:
[tex]N_1V_1=N_2V_2[/tex]
[tex]0.15\times22.2=N_2\times 9.45[/tex]
[tex]N_2=0.35[/tex]
[tex]Normality =molarity \times n-factor[/tex]
But in case of acid it is known as basicity and in case of base it is known as acidity.
in case of NaOH and HCl n-factor is 1;
hence
Normality=molarity;
[tex]molarity=0.35 mole/litre = final concentration of HCl[/tex]