a 0.20 mol/l solution of weak base hpo4 has a ph of 9. find the Kb
To find the Kb (base dissociation constant) of the weak base HPO4 (hydrogen phosphate), we need to use the pH of the solution and the balanced chemical equation for the dissociation of the base in water.
Here is the balanced chemical equation for the dissociation of HPO4:
HPO4- (aq) + H2O (l) ⇌ H2PO4- (aq) + OH- (aq)
From the equation, we can see that one hydroxide ion (OH-) is produced for every molecule of HPO4- that dissociates. Therefore, the concentration of OH- in the solution will be equal to the concentration of HPO4-.
Given that the solution has a pH of 9, we know that the concentration of H+ ions is 10^(-pH) = 10^(-9) M. Since the solution is basic, we can assume that the concentration of OH- is also 10^(-9) M.
Now, let's use the concentration of OH- to calculate the concentration of HPO4-:
OH- concentration = HPO4- concentration = 10^(-9) M
Next, we need the initial concentration of the weak base HPO4. It is given in the problem as 0.20 mol/L. However, we need to convert it into moles per liter of water for accurate calculations.
The total volume of the solution is not provided in the problem. If we assume that it is 1 liter, then the initial concentration of HPO4- is 0.20 mol/L.
Finally, we can use the concentrations of OH- and HPO4- to calculate Kb.
Kb = [OH-][HPO4-] / [H2PO4-]
Since HPO4- is the base and dissociates to form OH- and H2PO4-, we can assume that the concentration of H2PO4- is negligible compared to the initial concentration of HPO4-. Therefore, we can ignore the concentration of H2PO4- in the denominator.
Substituting the values:
Kb = (10^(-9) M) * (10^(-9) M) / (0.20 M)
Simplifying this expression gives us the value of Kb for the weak base HPO4.