Sodium benzoate. NaC_7H_5O_2, is used as a preservative in foods. Consider a 50.

Question

Sodium benzoate. NaC_7H_5O_2, is used as a preservative in foods. Consider a 50.0-mL sample of 0.250 M NaC_7H_5O_2 being titrated by 0.200 M HBr. Calculate the pH of the solution after the addition of 50.0 mL of the HBr solution? State whether 1 M solutions of the following salts in water are acidic, basic, or neutral. (modified text prob) a. K_2CO_3 b. NH_4C_2H_3O_2 c. Cu(NO_2)_2 d. KNO_3 e. CH_3NH_3Cl Consider the weak acids in the attached Ka table. Which acid-base pair would be best for a buffer at a pH of a. 4.5 b. 9.2 c.11.0

Explanation / Answer

16) Write down the balanced chemical reaction for the reaction between sodium benzoate, NaC7H5O2 and HBr as below:

NaC7H5O2 + HBr --------> NaBr + HC7H5O2

As per the balanced stoichiometric reaction,

1 mole NaC7H5O2 = 1 mole HBr.

Moles of NaC7H5O2 = (50.0 mL)*(0.250 mol/L) = 12.5 mmole.

Moles of HBr = (50.0 mL)*(0.200 mol/L) = 10.0 mmole.

Therefore, mmoles of NaC7H5O2 neutralized = 10.0 and mmoles of HC7H5O2 formed = 10.0.

Mmoles of NaC7H5O2 retained = (12.5 – 10.0) mmole = 2.5 mmole.

Total volume of the solution = (50.0 + 50.0) mL = 100.0 mL.

[NaC7H5O2] = (2.5 mmole)/(100.0 mL) = (2.5/100) M

[HC7H5O2] = (10.0 mmole)/(100.0 mL) = (10.0/100) M

Use the Henderson-Hasslebach equation as

pH = pKa + log [NaC7H5O2]/[HC7H5O2] = 4.20 + log [(2.5/100) M/(10.0/100) M] = 4.20 + log (2.5/10.0) = 4.20 + log (0.25) = 4.20 + (-0.602) = 3.598 (ans).

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