Question

what is the concentration of each substance if a solution is prepared by mixing 5.00g of NaC2H3O2 with 15.0 ml of 1.45M acetic acid and then bringing the total volume to 250.0 ml?

Please show work

Answer 1

moles of sodium acetate are (5/82), =0.061, or 0.061/0.25= 0.244M

Ka of acetic acid = 1.75 x 10^-5

sodium acetate will dissociate completely as it's a salt, while acetic acud will not dissociate completely.

HC₂H₃O₂(aq) + H₂O(l)      H₃O+(aq) + C₂H₃O₂^-(aq)

concentration of acid =

N1v1= N2V2

N2= 0.087M

c^2/(1-) = Ka

or C= 1.237x 10^-3 = concentration of dissociated acid and the concentraion of H⁺

also concentraion of acid generated due to hydrolysis of sodium acetate is

[H⁺] = (K_wK_a/C)

=1.74x10^-9 ( which is negligible)

concentration of acetate = salt +acid

0.244+1.237x10^-3

=0.245M of acetate

concentration of sodium ion = concentratio of salt = 0.244M

concentraio of actate = 0.245 M

concentratio of [H⁺]= 1.27x10^-3