Question

please write typed answer

If we have a 1.0M acetic acid solution what will be its PH? Ka=1.85x10-5 Write the dissociation (and hydrolysis if relevant) of each in water: ki CH3CO2Na NH4Cl

Answer 1

1 m CH3COOH

ka=1.8 *10^-5

pH=?

CH3COOH +H20 =H30+ + CH3COO-

INITIAL 1M 0 0

EQB 1-x x x

ka=(H3O+) (CH3COO-) / (CH3COOH)

1.8 *10^ -5 =x*x / 1-x *(1-x~ 1)

x^2 =1.8 * 10^-5

x=4.24 *10^-3

(H3O+) =4.24 *10^-3

pH=-log (H30+)

pH= 2.37

1 -kcl dissociation is

kcl(s) +H2O(l) =K+(aq) + cl-(aq)

2-sodium acetate CH3COONa

complete dissociation after dissolving

CH3COONa (s) - CH3COO-   + Na+

CH3COO- +H2O - CH3COOH + OH-

3- NH4Cl dissociation

NH4CL + H2O - NH4+ + CL-

thanx plz rate positive