Q2: What is the pH and concentration of CN– in a 0.100M solution of HCN? Q3:...

Q2: What is the pH and concentration of CN– in a 0.100M solution of HCN?

Q3: What would happen if the equilibrium [CN– ] that you calculated in Q2 was changed by adding 0.00500 moles of solid NaCN to 100.0 mL of the solution? (Which direction is the reaction going to “go” to minimize the disturbance?)

Expert Solution

Q 2.

HCN is a weak acid with K_a value of 6.2 * 10^-10

The concentration of of CN^- ion in equilibrium is : (Ka * [HCN])^1/2 = ( 6.2 * 10^-10 * 0.100) ^1/2 M = 7.87 * 10^-6 M

Concentration of H⁺ ion in equilibrium = concentration of of CN^- ion in equilibrium = 7.87 * 10^-6 M

pH = - log[H⁺] = - log (7.87 * 10^-6 ) = 5.10

Q 3.

If solid NaCN is added to the solution, then NaCN dissociates to produce CN^- ion in solution. As the concentration of CN- increases, to maintain equilibrium, the reaction shifts to reactant side i.e. more HCN produces.

queen_honey_blossom answered 2 years ago

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