Cyanide solutions used for heap leaching are made up of sodium cyanide (NaCN). T

Question

Cyanide solutions used for heap leaching are made up of sodium cyanide (NaCN). The sodium cyanide solution ([NaCN]-total = 0.01 M) is kept at pH 10.4. The pH is kept high to keep the cyanide in the CN- form. If CN protonates by the reverse of the HCN dissociation reaction,

HCN = H+ + CN-                   pKa = 9.24

and forms HCN, it will volatilize, and if it volatilizes, HCN can poison mine workers. At pH 10.4, what fraction of the cyanide is in the CN- form?

Calculate the fraction as [CN-] / [NaCN]-total.

Explanation / Answer

HCN <=> H+ + CN-

0 4*10^(-11) 0.01

At equilibrium,

x 4*10^(-11)-x 0.01-x

Ka = 5.75*10^(-10)

= (4*10^(-11)-x)(0.01-x)/x

Solving the equation, x = 0.11 or x= 4*10^(-11)

Hence, x= 4*10^(-11) is the only possible solution since x cannot take the value 0.11

[CN-] = 0.01-(4*10^(-11))

[NaCN]-total = 0.01

ratio = 0.999999996

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