Answer:
11.61 is the pH of 10.0 mL of a solution containing 3.96 g of sodium stearate.
Explanation:
Concentration of sodium stearate acid : c
Moles of sodium stearate = [tex]\frac{3.96 g}{306 g/mol}=0.01294 mol[/tex]
Volume of the solution = 10.0 mL = 0.010 L
[tex]c=\frac{0.01294 mol}{0.010 L}=1.294 M[/tex]
[tex]C_{17}H_{35}COONa\rightleftharpoons C_{17}H_{35}COO^-+Na^+[/tex]
[tex][C_{17}H_{35}COO^-]=c=1.294 M[/tex]
[tex]C_{17}H_{35}COO^-+H_2O\rightleftharpoons C_{17}H_{35}COOH +OH^-[/tex]
initially c
c 0 0
At equilibrium
(c-x) x x
Dissociation constant of an acid = [tex]K_a=1.3\times 10^{-5}[/tex]
Expression of a dissociation constant of an acid is given by:
[tex]K_a=\frac{[C_{17}H_{35}COOH][OH^-]}{[C_{17}H_{35}COO^-]}[/tex]
[tex]K_a=\frac{(x)^2\times x}{(c -x)}[/tex]
[tex]1.3\times 10^{-5}=\frac{x^2}{1.294-x}[/tex]
Solving for x;
x = 0.0041 M
[tex][OH^-]=0.0041 M[/tex]
The pOH of the solution:
[tex]pOH=-\log[OH^-]=-\log[0.0041 M]=2.39[/tex]
pH = 14 -pOH
pH = 14 - 2.39 = 11.61
11.61 is the pH of 10.0 mL of a solution containing 3.96 g of sodium stearate.
