Answer:
a: before equivalence point
b: equivalence point
c: before equivalence point
d: after the eqivalence point
e: before equivalence point
f: after the eqivalence point
Explanation:
Balanced equation of reaction:
NaOH +HCl =NaCl +H2O;
Volume of HCl is fixed and it 100ml and concentration is 1.0M
N1 and N2 normality of HCl and NaOH respectively;
V1 and V2 volume of HCl and NaOH respectively;
we have given molarity but we need normality;
![Normality=molarity \times n-factor](https://tex.z-dn.net/?f=Normality%3Dmolarity%20%5Ctimes%20n-factor)
<em>but in case of NaOH and HCl n-factor is 1 for each.</em>
hence
normality=molarity;
At equivalence point: ![N_1V_1=N_2V_2](https://tex.z-dn.net/?f=N_1V_1%3DN_2V_2)
Before equivalence point : ![N_1V_1>N_2V_2](https://tex.z-dn.net/?f=N_1V_1%3EN_2V_2)
After the equivalence point: ![N_1V_1](https://tex.z-dn.net/?f=N_1V_1%3CN_2V_2)
![N_1V_1=100\times1=100](https://tex.z-dn.net/?f=N_1V_1%3D100%5Ctimes1%3D100)
case a: 5.00 mL of 1.00 M NaOH
![N_2V_2=5\times1=5](https://tex.z-dn.net/?f=N_2V_2%3D5%5Ctimes1%3D5)
hence it is before equivalence point
case b: 100mL of 1.00 M NaOH
![N_2V_2=100\times1=100](https://tex.z-dn.net/?f=N_2V_2%3D100%5Ctimes1%3D100)
hence it is equivalence point
case c: 10.0 mL of 1.00 M NaOH
![N_2V_2=10\times1=10](https://tex.z-dn.net/?f=N_2V_2%3D10%5Ctimes1%3D10)
hence it is before equivalence point
case d: 150 mL of 1.00 M NaOH
![N_2V_2=150\times1=150](https://tex.z-dn.net/?f=N_2V_2%3D150%5Ctimes1%3D150)
hence it is after the eqivalence point
case e: 50.0 mL of 1.00 M NaOH
![N_2V_2=50\times1=50](https://tex.z-dn.net/?f=N_2V_2%3D50%5Ctimes1%3D50)
hence it is before equivalence point
case f: 200 mL of 1.00 M NaOH
![N_2V_2=200\times1=200](https://tex.z-dn.net/?f=N_2V_2%3D200%5Ctimes1%3D200)
hence it is after the eqivalence point