Sodium benzoate (C6H, CO₂Na) is wed as a food Preservative! Part A Calculate the pt in...

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Sodium benzoate (C6H, CO₂Na) is wed as a food Preservative! Part A Calculate the pt in 0.058 M sodium benzateo ka for benzoic

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Answer 1

Answer #1

part-A

C6H5CO2Na(aq) ------------> C6H5CO2^- (aq) + Na^+ (aq)

0.058M --------------------------- 0.058M ---------------- 0.058M

[Na^+] = 0.058M

Kb = Kw/Ka

= 10^-14/(6.5*10^-5) = 1.54*10^-10

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

I ----------- 0.058 ------------------------------------------------ 0 ---------------------- 0

C------------ -x ------------------------------------------------- +x --------------------- +x

E---------- 0.058-x ---------------------------------------------- +x -------------------- +x

Kb = [C6H5CO2H][OH^-]/[C6H5O2^-]

1.54*10^-10 = x*x/(0.058-x)

1.54*10^-10(0.058-x) = x^2

x = 3*10^-6

[C6H5CO2H] = x = 3*10^-6M

[OH^-] = x = 3*10^-6M

[C6H5CO2^-] = 0.058-x =0.058-3*10^-6 = 0.057997M

[H3O^+] = Kw/[OH^-]

= 1*10^-14/(3*10^-6)

= 3.34*10^-9 M

PH = -log[H^+]

= -log(3.34*10^-9)

= 8.4762

part-B

6H5CO2Na(aq) ------------> C6H5CO2^- (aq) + Na^+ (aq)

0.058M --------------------------- 0.058M ---------------- 0.058M

[Na^+] = 0.058M

Kb = Kw/Ka

= 10^-14/(6.5*10^-5) = 1.54*10^-10

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

I ----------- 0.058 ------------------------------------------------ 0 ---------------------- 0

C------------ -x ------------------------------------------------- +x --------------------- +x

E---------- 0.058-x ---------------------------------------------- +x -------------------- +x

Kb = [C6H5CO2H][OH^-]/[C6H5O2^-]

1.54*10^-10 = x*x/(0.058-x)

1.54*10^-10(0.058-x) = x^2

x = 3*10^-6

[C6H5CO2H] = x = 3*10^-6M

[OH^-] = x = 3*10^-6M

[C6H5CO2^-] = 0.058-x =0.058-3*10^-6 = 0.057997M

[H3O^+] = Kw/[OH^-]

= 1*10^-14/(3*10^-6)

= 3.34*10^-9 M

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