Solve Ka for Weak Acid HCN - 0.18mol KCN in 1.00L, pH 9.70, 298K

[Aly]

hi forum,
would nebody be able to help me solve this problem?

0.18mol of potassium cyanide (KCN) was dissolved in 1.00L of a solution in which the pH was held constant at 9.70 at a temperature of 298K.

the equilibrium concentration of CN- was 0.13M.

Calculate Ka for the weak acid HCN.

ne help would be greatly appreciated. :)

 

[Borek]

Use Henderson-Hassebalch equation. It won't be difficult to calculate equilibrium concentrations of HCN and CN- - just enter them into HH and solve for pKa.

 

[GCT]

hi forum,
would nebody be able to help me solve this problem?

0.18mol of potassium cyanide (KCN) was dissolved in 1.00L of a solution in which the pH was held constant at 9.70 at a temperature of 298K.

the equilibrium concentration of CN- was 0.13M.

Calculate Ka for the weak acid HCN.

ne help would be greatly appreciated. :)

Start with finding Kb

Kb= \frac{[OH-][HCN]}{[CN-]}

You can find the initial concentration, you also know the equilibrium concentration, thus you can find how much of CN- reacted. It also says that the pH was held constant. From the pH given, find the pOH, from this calculate the concentration of hydroxide [OH-]. This is the equilibrium concentration thus you can plug it back into the equilibrium equation.

How would you find the HCN concentration to solve for Kb...and then Ka?