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

Question

15114 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-20T07:55:13

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

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

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

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.