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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

Respuesta :

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;

[tex]Normality=molarity \times n-factor[/tex]

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

hence

normality=molarity;

At equivalence point:  [tex]N_1V_1=N_2V_2[/tex]

Before equivalence point : [tex]N_1V_1>N_2V_2[/tex]

After the equivalence point: [tex]N_1V_1<N_2V_2[/tex]

[tex]N_1V_1=100\times1=100[/tex]

case a:  5.00 mL of 1.00 M NaOH

[tex]N_2V_2=5\times1=5[/tex]

[tex]N_1V_1>N_2V_2[/tex] hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

[tex]N_2V_2=100\times1=100[/tex]

[tex]N_1V_1=N_2V_2[/tex] hence it is equivalence point

case c:  10.0 mL of 1.00 M NaOH

[tex]N_2V_2=10\times1=10[/tex]

[tex]N_1V_1>N_2V_2[/tex] hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

[tex]N_2V_2=150\times1=150[/tex]

[tex]N_1V_1<N_2V_2[/tex] hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

[tex]N_2V_2=50\times1=50[/tex]

[tex]N_1V_1>N_2V_2[/tex] hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

[tex]N_2V_2=200\times1=200[/tex]

[tex]N_1V_1<N_2V_2[/tex] hence it is after the eqivalence point

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