calculate the pH pOH and (OH(aq)) of 0.006mol/l HI(aq)
HI is a strong acid; i.e., it ionizes 100%; therefore, (H^+) = 0.006M
pH = -log(H^+)
Get pOH from this.
pH + pOH = pKw = 14
Then OH^- from pOH = -log(OH^-)
To calculate the pH, pOH, and the concentration of hydroxide ions ([OH^-]) in a solution of HI(aq), we need to consider the dissociation of HI in water.
The balanced chemical equation for the dissociation of HI(aq) is:
HI(aq) → H^+(aq) + I^-(aq)
From this equation, we can see that one mole of HI dissociates to produce one mole of H^+ ions and one mole of I^- ions.
Since the concentration of HI is given as 0.006 mol/L, the concentration of H^+ ions will also be 0.006 mol/L.
pH is a measure of the concentration of H^+ ions in a solution. It is calculated using the equation:
pH = -log [H^+]
Therefore, in this case, we have:
pH = -log (0.006)
= -log (6 × 10^-3)
= -(-2.22) (taking log (6 × 10^-3) = -2.22)
= 2.22
Therefore, the pH of the solution is 2.22.
pOH is a measure of the concentration of OH^- ions in a solution. Since the solution is acidic (because it contains H^+ ions), the concentration of OH^- ions can be considered negligible, and thus the pOH can be assumed to be 0.
The relationship between pH and pOH is given by the equation:
pH + pOH = 14
Therefore, if the pH is 2.22, the pOH can be calculated as:
pOH = 14 - pH
= 14 - 2.22
= 11.78
Therefore, the pOH of the solution is 11.78.
Since the solution is acidic, the concentration of OH^- ions can be assumed to be negligible. Therefore, [OH^-] is very close to 0 in this case.