Question

Sodium benzoate, NaC7H502 (the numbers here are subscripts) , is used as a preservative in foods. Consider a 50.0-mL sample of 0.250 MNaC7H502 being titrated by 0.200 M HBr. Calculate the pH of the solution: (A) when no HBr has been added; (B) after the addition of 50.0 ml-a of the HBr solution; (C) at the equivalence point; (D) after the addition of 75.00 mL of the HBr solution. The -10Kb value for the benzoate ion is 1.6 X 10 to the -10th power.

Answer 1

Answer on Question #59840, Chemistry / General Chemistry

Sodium benzoate, $\mathrm{NaC_7H_5O_2}$, is used as a preservative in foods. Consider a 50.0-mL sample of 0.250 M $\mathrm{NaC_7H_5O_2}$ being titrated by 0.200 M HBr. Calculate the pH of the solution: (A) when no HBr has been added; (B) after the addition of 50.0 mL of the HBr solution; (C) at the equivalence point; (D) after the addition of 75.00 mL of the HBr solution. The -10Kb value for the benzoate ion is 1.6 X 10 to the -10th power.

Solution:

A) Sodium benzoate is the salt formed by the strong basis of NAON and weak C6H5COOH acid which in water solution is hydrolyzed on anion:

$$\mathrm{C_6H_5COONa} + \mathrm{HOH} \leftrightarrow \mathrm{NaOH} + \mathrm{C_6H_5COOH}$$

The constant of dissociation of benzoic acid is equal to $K = 6,3 \cdot 10^{-5}$, $pK = 4.20$

pH of initial solution of sodium benzoate we will calculate on a formula:

$$\mathrm{pH} = 7 + \frac{1}{2} \cdot \mathrm{pK} + \frac{1}{2} (\lg C_{\text{salt}}) = 7 + \frac{1}{2} \cdot 4.20 + \frac{1}{2} \lg (0,250) = 8.980$$

B) Titration of benzoate of sodium happens on the equation:

$$\mathrm{C_6H_5COONa} + \mathrm{HBr} \leftrightarrow \mathrm{NaBr} + \mathrm{C_6H_5COOH}$$

Thus, in titrable solution there is a buffer solution – mix of weak, benzoic acid with her salt. pH such buffer mix we will calculate value on a formula:

$$\mathrm{pH} = \mathrm{pK_a} - \lg \left( \frac{C_{\text{acid}}}{C_{\text{salt}}} \right)$$

As both concentration pay off in the same volume, in this formula $C_{\text{acid}} / C_{\text{salt}} = n_{\text{acid}} / n_{\text{salt}}$ is possible.

$$n_{\text{salt}} = M_{\text{salt}}^0 \cdot V_{\text{salt}}^0 - M_{\text{HBr}} \cdot V_{\text{HBr}}.$$

Then pH it is possible to calculate on a formula:

$$\mathrm{pH} = 4.20 - \lg \left( \frac{V_{\text{HBr}} \cdot 0.200}{0.250 V_{\text{salt}}^0 - 0.200 V_{\text{HBr}}} \right)$$

After addition of 50 ml of HBr solution:

$$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} - \lg \left(\frac{50 \cdot 0.200}{50 \cdot 0.250 - 50 \cdot 0.200}\right) = 4.20 - (-1,10) = 5,30$$

C) In an equivalence point all sodium benzoate has already passed into benzoic acid. We will calculate the volume of the added solution of acid on a formula:

$$M_{\text{salt}}^{\mathrm{O}}\cdot V_{\text{salt}}^{\mathrm{O}} = M_{\mathrm{HBr}}\cdot V_{\mathrm{HBr}},$$ $$\text{Then } V_{\mathrm{HBr}} = M_{\text{salt}}^{\mathrm{O}}\cdot V_{\text{salt}}^{\mathrm{O}} / M_{\mathrm{HBr}} = 0,250 \cdot 50 / 0,200 = 62,5 \text{ ml}$$

Solution volume is equal in a point of equivalence $50 + 62,5 = 112,5$ ml therefore concentration of benzoic acid is equal $0,200 \cdot 62,5 / 112,5 = 0,111$ M. We will calculate dissociation degree

$$K = \frac{C^2 \alpha^2}{C(1 - \alpha)} = \frac{C \alpha^2}{(1 - \alpha)} \text{ Since } \alpha \ll 1, \text{ then } (1 - \alpha) \approx 1.$$ $$\text{then } \alpha = \sqrt{\frac{K}{C}} = \sqrt{\frac{6.3 \times 10^{-5}}{0.111}} = 2.38 \times 10^{-2}$$

Then concentration of ions of hydrogen is equal

$$[H^+] = 0.111 \cdot 2.38 \times 10^{-2} = 2.64 \times 10^{-3}$$ $$\mathrm{pH} = -\lg [H^+] = 2,58$$

D) After an equivalence point pH solution will be defined by excess concentration of HBr. As Br – strong acid, then pH = – lg[HBr]

After addition of 75 ml of HBr solution excess concentration

$$[HBr] = (75 - 62.5) \cdot 0.2000 / (50 + 75) = 0.02 \text{ M}$$ $$\text{Then } pH = -\lg(0.02) = 1.70$$

http://www.AssignmentExpert.com/