Answer to Question #107416 in General Chemistry for kim

Question

1) If 10.0 mL of glacial acetic acid (pure HC2H3O2) is diluted to 1.50 L with water, what is the pH of the resulting solution? The density of glacial acetic acid is 1.05 g/mL.

2) A 0.190 M solution of a weak acid (HA) has a pH of 2.92. Calculate the acid ionization constant (Ka) for the acid.

3) Determine the percent ionization of a 0.250 M solution of benzoic acid.

Answer 1

1) glacial acetic acid is 100% acetic acid.

m(CH₃COOH) = 10*1.05 = 10.5 g

n(CH₃COOH) = ${\frac {10.5} {60}}=0.175$ mol

C(CH₃COOH) =${\frac {0.175} {1.5}}$ = 0.1167 M

x M of acid had dissociated:

0.1167-x x x

CH₃COOH = CH₃COO^- + H⁺

if Ka(CH₃COOH) = 1.74*10^-5 then

Ka = ${\frac {[H^+][CH_3COO^-]} {[CH_3COOH]}}={\frac {x^2} {0.1167-x}}=1.74*10^{-5}$

we neglect "-x" in denominator and resulting equation is

${\frac {x^2} {0.1167}}=1.74*10^{-5}$

x² = 1.74*10^-5*0.1167 = 2.03*10^-6

x =[H⁺]= 1.42*10^-3

pH = -lg[H+] = -lg(1.42*10^-3)=2.85

2)

HA = H⁺ + A^-

if x M of HA had dissociated then

[H⁺] = [A^-] = x

[HA] = 0.190 - x

Ka = ${\frac {[H^+][A^-]} {[HA]}}={\frac {[x]^2} {(0.190-x)}}$

[H+] = x = 10^-pH = 10^-2.92 = 0.0012 M

then Ka = ${\frac {(0.0012)^2} {(0.190-0.0012)}}=7.63*10^{-6}$

3)

Ka(C₆H₅COOH) = 6.17*10^-5

if x M of benzoic acid had dissociated then

0.250 - x x x

C₆H₅COOH = C₆H₅COO^- + H⁺

Ka = ${\frac {[H^+][C_6H_5COO^-]} {[C6H5COOH]}}={\frac {x^2} {0.250-x}}=6.17*10^{-5}$

After solving the equation we get

x = 3.9*10^-3 M

The percent ionization is ${\frac {3.9*10^{-3}} {0.25}}=1.56$ %