Question

Sodium benzoate (C6H5CO2Na) is used as a food preservative.

Part A

Calculate the pH in 0.052 M sodium benzoate; Ka for benzoic acid (C6H5CO2H) is 6.5×10−5.

Part B

Calculate the concentrations of all species present (Na+, C6H5CO−2, C6H5CO2H, H3O+ and OH−) in 0.052 M sodium benzoate.

Express your answers using two significant figures. Enter your answers numerically separated by commas.

Please show work thank you!

Answer 1

[C6H5CO2Na] = [ C6H5CO2- ]      …( salt dissociates completely)

Kb of benzoate ion = 1.0E-14/ ka ( benzoic acid) = 1.0E-14/ 6.5E-5 = 1.54E-10

Write the reaction of benzoate ion with water and set up ICE

            C6H5CO2- (aq)   + H2O (l) ----- > C6H5CO2H (aq)    +   OH- (aq)

I         0.052                                                      0                                 0

C     - x                                                            +x                                +x

E     (0.052-x)                                                  x                                  x

Use kb expression and find out x ( OH- concentration )

Kb = 1.54 E-10 = x² / (0.052 – x )

Since the value of kb is very small, we can neglect the value of x in the denominator.

X^2 = 1.54 E-10 x 0.052

x = 2.82 E-6

x =C6H5CO2H= [OH-] = 2.8 E-6 M

pOH = - log ([OH-]) = - log ( 2.82 E-6)

= 5.55

pH = 14 – pOH = 14 – 5.55 = 8.45 = 8.5

[C6H5CO2-] = 0.052 – x = 0.052 – 2.82 E-6 = 0.05199 M = 0.052 M

[Na+] = [CH3COONa] = 0.052 M