What is the pH of a 1.24 mol/L solution of HCN(aq) if its Ka = 6.2 x 10-10
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Write the equation.
HCN ==> H^+ + CN^-
Write the Ka expression.
Ka = (H^+)(CN^-)/(HCN) = 6.32 x 10^-10
Set up the ICE table and go from there.
To find the pH of a solution of HCN(aq) with a known concentration and Ka value, you can use the Henderson-Hasselbalch equation. This equation relates the pH of a solution to the concentration of the acid and its dissociation constant.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where pH is the logarithmic measure of hydrogen ion concentration, pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base (in this case, CN-) and [HA] is the concentration of the acid (HCN).
In this case, we're given the concentration of HCN(aq) as 1.24 mol/L and the Ka value as 6.2 x 10^-10.
First, let's convert the concentration of HCN into [HA]. Since HCN is a weak acid, we can assume that most of it will remain undissociated and contribute to the [HA] concentration. Therefore, [HA] = 1.24 mol/L.
Next, we can calculate the concentration of [A-] by using the equation:
[A-] = √(Ka × [HA])
Plugging in the values we have:
[A-] = √((6.2 x 10^-10) × (1.24))
≈ 7.96 x 10^-6 mol/L
Now, we can substitute these values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
= -log(6.2 x 10^-10) + log((7.96 x 10^-6)/(1.24))
≈ -9.208 + 2.06
≈ -7.147
Therefore, the pH of the 1.24 mol/L solution of HCN(aq) is approximately 7.15.