How do you calculate the pH of a 0.050 M sodium cyanide solution?

1 Answer
May 20, 2016

Answer:

#pOH = pK_w-pH# #=# #14-3#.

Explanation:

#""^(-)C-=N(aq) + H_2O rarr H-C-=N + HO^-#

#K_b=([HC-=N][HO^-])/([""^(-)C-=N])# #=# #2.1xx10^-5#

Let the concentration of cyanide that associates #=# #x#, and given this stoichiometry:

#x^2/(0.05-x)# #=# #2.1xx10^-5#

This is a quadratic in #x#, which is solvable, but we can make the reasonable assumption that #(0.05-x)~=0.05#

#x# #~=# #sqrt{2.1xx10^-5xx0.05}# #=# #0.001#

Thus #[HO^-]# #=# #0.001*mol*L^-1#, #pOH# #=# #3#, and #pH# #=# #11#

Successive approximations would not markedly change ths value. Our assumption that #(0.05-x)~=0.05# was entirely justified.