Hydrogen cyanide accidentally gets added to a water storage bladder on an Army b

Question

Hydrogen cyanide accidentally gets added to a water storage bladder on an Army base. If the equilibrium pH is 6.15, calculate [CN], [HCN], and [OH]. (Hint: Write the equilibrium equations and the charge balance.)

3. Hydrogen cyanide (HCN) is a toxic chemical and is regulated in drinking water at a concentration of 0.2 mg/L (CN). Listed below are reactions relevant to HCN in water. Hydrogen cyanide accidentally gets added to a water storage bladder on an Army base. If the equilibrium pH is 6.15, calculate [CN], [HCN], and [OH] (Hint: Write the equilibrium equations and the charge balance.)

Explanation / Answer

Answer

[HCN] = 9.61×10-4M

[CN-]= 6.94×10-7M

[OH-] = 1.41×10-8M

Explanation

HCN(aq) <- - - - - > H+  + CN-

H2O(aq) < - - - - - - > H+ + OH-

To get charge balance

[H+] = [OH-] + [CN-]

so,

[CN-] = [H+] - [OH-]

given pH = 6.15

pOH = 14 - pH = 14 - 6.15 = 7.85

pH = - log[H+]

- log[H+] = 6.15

[H+] = 7.08×10-7M

pOH = - log[OH-]

-log[OH-] = 7.85

[OH-] = 1.41×10-8M

Therefore,

[CN-] = 7.08 ×10-7M - 1.41×10-8 = 6.94×10-7M

pKa of HCN = 9.3

   pKa = - logKa

   - logKa = 9.3

   Ka = 5.01×10-10

HCN <------> H+ + CN-

Ka = [CN-][H+] /[HCN] = 5.01×10-10

[CN-] = [H+]

Therefore,

[HCN] = (6.94×10-7)2/5.01×10-10 = 9.61×10-4 M   

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