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melamori03 [73]
3 years ago
12

Consider the titration of a 20.0-mL sample of 0.105 M HC2H3O2 with 0.125 M NaOH. Determine each quantity. a. the initial pH b. t

he volume of added base required to reach the equivalence point c. the pH at 5.0 mL of added base d. the pH at one-half of the equivalence point e. the pH at the equivalence point f. the pH after adding 5.0 mL of base beyond the equivalence point
Chemistry
1 answer:
Oksi-84 [34.3K]3 years ago
3 0

Answer:

Explanation:

Given that:

Concentration of HC_2H_3O_2 \  (M_1) = 0.105 M

Volume of  HC_2H_3O_2 \  (V_1) = 20.0 mL

Concentration of NaOH (M_2) = 0.125 M

The  chemical reaction can be expressed as:

HC_2H_3O_2_{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O_{(l)}

Using the ICE Table to determine the equilibrium concentrations.

          HC_2 H_3 O_2 _{(aq)} + H_2O _{(l) } \to C_2 H_3O_2^- _{(aq)} + H_3O^+_{ (aq)}

I            0.105                                     0                  0

C              -x                                         +x                +x

E            0.105 - x                                  x                  x

K_a = \dfrac{[C_2H_5O^-_2][H_3O^+]}{[HC_2H_3O_2]}

K_a = \dfrac{(x)(x)}{(0.105-x)}

Recall that the ka for HC_2H_3O_2= 1.8 \times 10^{-5}

Then;

1.8 \times 10^{-5} = \dfrac{(x)(x)}{(0.105 -x)}

1.8 \times 10^{-5} = \dfrac{x^2}{(0.105 -x)}

By solving the above mathematical expression;

x = 0.00137 M

H_3O^+ = x = 0.00137  \ M \\ \\  pH = - log [H_3O^+]  \\ \\  pH = - log ( 0.00137 )

pH = 2.86

Hence, the initial pH = 2.86

b)  To determine the volume of the added base needed to reach the equivalence point by using the formula:

M_1 V_1 = M_2 V_2

V_2= \dfrac{M_1V_1}{M_2}

V_2= \dfrac{0.105 \ M \times 20.0 \ mL }{0.125 \ M}

V_2 = 16.8 mL

Thus, the volume of the added base needed to reach the equivalence point = 16.8 mL

c) when pH of 5.0 mL of the base is added.

The Initial moles of HC_2H_3O_2 = molarity × volume

= 0.105  \ M \times 20.0 \times 10^{-3} \ L

= 2.1 \times 10^{-3}

number of moles of 5.0 NaOH = molarity × volume

number of moles of 5.0 NaOH = 0.625 \times 10^{-3}

After reacting with 5.0 mL NaOH, the number of moles is as follows:

                    HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}

Initial moles   2.1*10^{-3}       0.625 * 10^{-3}           0                      0

F(moles) (2.1*10^{-3} - 0.625 \times 10^{-3})    0      0.625 \times 10^{-3}         0.625 \times 10^{-3}

The pH of the solution is then calculated as follows:

pH = pKa + log \dfrac{[base]} {[acid]}

Recall that:

pKa for HC_2H_3O_2=4.74

Then; we replace the concentration with the number of moles since the volume of acid and base are equal

∴

pH = 4.74 + log \dfrac{0.625 \times 10^{-3}}{1.475 \times 10^{-3}}

pH = 4.37

Thus, the pH of the solution after the addition of 5.0 mL of NaOH = 4.37

d)

We need to understand that the pH at 1/2 of the equivalence point is equal to the concentration of the base and the acid.

Therefore;

pH = pKa = 4.74

e) pH at the equivalence point.

Here, the pH of the solution is the result of the reaction in the (C_2H_3O^-_2) with H_2O

The total volume(V) of the solution = V(acid) + V(of the base added to reach equivalence point)

The total volume(V) of the solution = 20.0 mL + 16.8 mL

The total volume(V) of the solution = 36.8 mL

Concentration of (C_2H_3O^-_2) = moles/volume

= \dfrac{2.1 \times 10^{-3} \ moles}{0.0368 \ L}

= 0.0571 M

Now, using the ICE table to determine the concentration of H_3O^+;

             C_2H_5O^-_2 _{(aq)} + H_2O_{(l)} \to HC_2H_3O_2_{(aq)} + OH^-_{(aq)}

I              0.0571                                0                      0

C              -x                                       +x                     +x

E             0.0571 - x                             x                       x

Recall that the Ka for HC_2H_3O_2 = 1.8 \times 10^{-5}

K_b = \dfrac{K_w}{K_a} = \dfrac{1.0\times 10^{-14}}{1.8 \times 10^{-5} }  \\ \\ K_b = 5.6 \times 10^{-10}

k_b = \dfrac{[ HC_2H_3O_2] [OH^-]}{[C_2H_3O^-_2]}

5.6 \times 10^{-10} = \dfrac{x *x }{0.0571 -x}

x = [OH^-] = 5.6 \times 10^{-6} \ M

[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{5.6 \times 10^{-6} }

[H_3O^+] =1.77 \times 10^{-9}

pH =-log  [H_3O^+]   \\ \\  pH =-log (1.77 \times 10^{-9}) \\ \\ \mathbf{pH = 8.75 }

Hence, the pH of the solution at equivalence point = 8.75

f) The pH after 5.09 mL base is added beyond (E) point.

             HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}

Before                             0.0021              0.002725         0

After                                   0                     0.000625        0.0021

[OH^-] = \dfrac{0.000625 \ moles}{(0.02 + 0.0218 )  \ L}

[OH^-] = \dfrac{0.000625 \ moles}{0.0418 \ L}

[OH^-] =  0.0149 \ M

From above; we can determine the concentration of H_3O^+ by using the following method:

[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{0.0149}

[H_3O^+] = 6.7 \times 10^{-13}

pH = - log [H_3O^+]

pH = -log (6.7 \times 10^{-13} )

pH = 12.17

Finally, the pH of the solution after adding 5.0 mL of NaOH beyond (E) point = 12.17

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