A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to the reaction flask. Classify the following conditions based on whether they are before the equivalence point, at the equivalence point, or after the equivalence point.

a. 5.00 mL of 1.00 M NaOH
b. 100mL of 1.00 M NaOH
c. 10.0 mL of 1.00 M NaOH
d. 150 mL of 1.00 M NaOH
e. 50.0 mL of 1.00 M NaOH
f. 200 mL of 1.00 M NaOH

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;

but in case of NaOH and HCl n-factor is 1 for each.

hence

normality=molarity;

At equivalence point:  

Before equivalence point :

After the equivalence point:

case a:  5.00 mL of 1.00 M NaOH

hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

hence it is equivalence point

case c:  10.0 mL of 1.00 M NaOH

hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

hence it is after the eqivalence point

electron microscopy and scanning tunneling microscopes gives us a greater understanding of atoms by being able to see the atom itself, and lets us see some atomic reactions.

explanation:

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.