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3 years ago

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Consider the titration of 100.0 mL of 0.100 M acetic acid with 0.100 M NaOH. CH3CO2H(aq) + OH-(aq) → CH3CO2-(aq) + H2O(ℓ) Ka for

acetic acid = 1.8×10-5 Kb for acetate ion = 5.6×10-10 (a) What is the pH of the solution when 90.0 mL of 0.100 M NaOH has been added to 100.0 mL of 0.100 M acetic acid? (b) What is the pH at the equivalence point? (c) What is the pH after 110.0 mL of NaOH has been added?

1 answer:

allsm [11]3 years ago

4 0

Answer:

a) pH = 5.70

b) pH = 8.22

c)pH = 11.68

Explanation:

a)NaOH (aq)+CH3COOH(aq) ------> CH3COONa(aq) + H2O(aq)

1mol of NaOH react with 1mol of CH3COOH

No of mole in 90ml of 0.1M NaOH

= (0.1mol/1000ml)×90ml

= 0.009

No of mol in 100ml of CH3COOH

= (0.1/1000)×100

= 0.01

No of mol of CH3COOH after addition

= 0.01-0.009

= 0.001

Total volume = 100ml + 90ml

= 190ml

Final molarity of CH3COOH =( 0.0025/190) ×1000

= 0.00526M

Concentration of CH3COONa formed =( 0.0075/190) ×1000

= 0.0474M

Ka of CH3COOH = 1.8 × 10^-5

pka = -log(ka)

pKa = 4.75

Applying Henderson equation

pH = pKa + log ( [A-]/[HA])

pH = 4.75 + log ( 0.0474/0.00526)

= 5.70

b)

At equivalencepoint point ,

No of moles of CH3COOH = 0

No of moles of CH3COO- = 0.01 mol

Total volume = 200ml

molarity of CH3COO- = 0.01/2

= 0.0050M

CH3COO- (aq) + H2O(l) <---------> CH3COOH(aq) + OH-(aq)

Kb = [ CH3COOH] [ OH- ] / [ CH3COO- ]

= 1.8 × 10^-5

[ CH3COOH ] = X

[ OH-] = X

[ CH3COO-] = 0.0050 - X

5.6 × 10^-10 = X^2/ (0.0050 - X)

we can assume , 0.0050 - X = 0.0050

5.6 × 10^-10 = X^2/0.0050

X = 1.67 × 10^6

[OH-] = 1.67 × 10^-6

pOH = 5.78

pH = 14 - pOH

pH = 14 -5.78

pH = 8.22

c) No of mol of OH from excess 10ml of NaOH = (0.1mol /1000ml)×10ml = 0.001mol

No of mol of OH- from hydrolysis of CH3COO- = (1.67×10^-6/1000)×200= 3.34×10^-7mol

Second one is negligible

So, no of mol OH- = 0.001mol

Total volume = 100ml + 110ml

= 210ml

[OH-] =( 0.001/210)×1000

= 0.0048M

pOH = -log[OH-]

= - log (0.0048)

= 2.32

pH = 14 - 2.32

pH = 11.68

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Refer to a modern periodic table for the relative atomic mass data:

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$M(\rm N_2O_4) = 14.007 \times 2 + 15.999 \times 4 = \rm 92.010\; g\cdot mol^{-1}$ .

$M(\rm NO_2) = 14.007 + 2\times 15.999 = \rm 46.005\; g\cdot mol^{-1}$ .

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$\displaystyle n(\mathrm{N_2O_4}) = \frac{m(\mathrm{N_2O_4}) }{M(\mathrm{N_2O_4})} = \rm 0.543419\; mol$ .

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$\displaystyle c = \frac{n}{V} = \rm 0.271710\; mol\cdot L^{-1}$ .

Construct a RICE table for this equilibrium. Note that all values in this table shall stand for concentrations. Let the change in the concentration of $\rm N_2O_4\; (g)$ be $-x\; \rm mol\cdot L^{-1}$ .

$\begin{array}{c|ccc}\mathbf{R} & \mathrm{N_2O_4}\; (g) & \rightleftharpoons & 2\;\mathrm{NO_2}\;(g)\\\mathbf{I} & 0.271710 & & \\ \mathbf{C} & -x & & +2\;x \\\mathbf{E} & 0.271710 -x & & 2x\end{array}$ .

The equilibrium concentrations shall satisfy the equilibrium law for this reaction under this particular temperature.

$\displaystyle \frac{[\rm NO_2]^{2}}{[\rm N_2O_4]} = \rm K_{c} = 0.133$ .

$\displaystyle \frac{(2x)^{2}}{0.271710 - x} = 0.133$ .

This equation can be simplified to a quadratic equation. Solve this equation. Note that there might be more than one possible values for $x$ . $x$ itself might not necessarily be positive. However, keep in mind that all concentrations in an equilibrium should be positive. Apply that property to check the $x$ -value.

$x \approx 0.0798672$ .

$\rm [N_2O_4] \approx 0.271710 - 0.0798672 = \rm 0.191843 \;mol\cdot L^{-1}$ .

$\rm [NO_2] \approx 2 \times 0.0798672 = 0.159734\; mol\cdot L^{-1}$ .

What will be the concentration of that additional $\rm 5.00\; g$ of $\rm[NO_2]$ if it was added to an evacuated $\rm 2.0\; L$ flask?

$\displaystyle n(\mathrm{NO_2}) = \frac{m(\mathrm{NO_2}) }{M(\mathrm{NO_2})} = \rm 1.08684\; mol$ .

$\displaystyle c = \frac{n}{V} = \frac{1.08684}{2.0} = \rm 0.543419\;mol\cdot L^{-1}$ .

The new concentration of $\mathrm{NO_2}\; (g)$ will become

$0.543419 + 0.159734 = \rm 0.703154\; mol\cdot L^{-1}$ .

Construct another RICE table. Let the change in the concentration of $\mathrm{N_2O_4}\; (g)$ be $-x\;\rm mol\cdot L^{-1}$ .

$\begin{array}{c|ccc}\mathbf{R} & \mathrm{N_2O_4}\; (g) & \rightleftharpoons & 2\;\mathrm{NO_2}\;(g)\\\mathbf{I} & 0.191843 & & 0.703154 \\ \mathbf{C} & -x & & +2\;x \\\mathbf{E} & 0.191843 -x & & 0.703154 + 2x\end{array}$ .

Once again, the equilibrium conditions shall satisfy this particular equilibrium law.

$\displaystyle \frac{[\rm NO_2]^{2}}{[\rm N_2O_4]} = \rm K_{c} = 0.133$

$\displaystyle \frac{(0.703154 + 2x)^{2}}{0.191843 -x} = 0.133$ .

Simplify and solve this equation for $x$ . Make sure that the $x$ -value ensures that all concentrations are positive.

$x \approx \rm -0.232758$ .

Note that $x \neq -0.503646$ for if that would lead to a negative value for the concentration of $\mathrm{NO}_2$ .

Hence the equilibrium concentration of $\rm N_2O_4$ will be:

$[{\rm N_2O_4}] = 0.703154 + 2(-0.232758) = \rm 0.237638\; mol\cdot L^{-1}$ .

$n = c\cdot V = 0.237638\times 2 = \rm 0.475276\; mol\cdot L^{-1}$ .

$m = n \cdot M = 0.475276 \times 92.010 \approx \rm 43.7\; g$

6 0

3 years ago