5. A solution of silver nitrate with a molar concentration of 0.1000mol/L is titrated by a solution of potassium thiocyanate with the same concentration. Calculate pAg, if to 10.00 ml of silver nitrate solution were added a) 9.90 ml; b) 10.00 ml and c) 10.1 ml of titrant.

To calculate the pAg of the solution, we need to determine the concentration of silver ions (Ag+) in the solution.<br /><br />Given information:<br />- The molar concentration of the silver nitrate (AgNO3) solution is 0.1000 mol/L.<br />- The molar concentration of the potassium thiocyanate (KSCN) solution is also 0.1000 mol/L.<br /><br />a) When 9.90 ml of KSCN solution is added to 10.00 ml of AgNO3 solution:<br />- The total volume of the solution becomes 19.90 ml.<br />- The concentration of Ag+ remains 0.1000 mol/L, as the volume of the AgNO3 solution is 10.00 ml.<br />- The concentration of SCN- becomes 0.1000 mol/L, as the volume of the KSCN solution is 9.90 ml.<br /><br />The reaction between Ag+ and SCN- is:<br />Ag+ + SCN- ⇌ [AgSCN]<br /><br />The equilibrium constant (K) for this reaction is:<br />K = [AgSCN] / ([Ag+] × [SCN-])<br /><br />Since the concentrations of Ag+ and SCN- are equal, the equilibrium constant can be simplified to:<br />K = 1<br /><br />The pAg is defined as the negative logarithm of the concentration of Ag+:<br />pAg = -log[Ag+]<br /><br />Since the concentration of Ag+ is 0.1000 mol/L, the pAg is:<br />pAg = -log(0.1000) = 1<br /><br />b) When 10.00 ml of KSCN solution is added to 10.00 ml of AgNO3 solution:<br />- The total volume of the solution becomes 20.00 ml.<br />- The concentration of Ag+ becomes 0.0500 mol/L, as the volume of the AgNO3 solution is halved.<br />- The concentration of SCN- becomes 0.1000 mol/L, as the volume of the KSCN solution is 10.00 ml.<br /><br />Using the same equilibrium constant (K = 1) as in part a), the pAg is:<br />pAg = -log(0.0500) = 1.3<br /><br />c) When 10.1 ml of KSCN solution is added to 10.00 ml of AgNO3 solution:<br /> solution becomes 20.10 ml.<br />- The concentration of Ag+ remains 0.1000 mol/L, as the volume of the AgNO3 solution is 10.00 ml.<br />- The concentration of SCN- becomes 0.0980 mol/L, as the volume of the KSCN solution is 10.1 ml.<br /><br />Using the same equilibrium constant (K = 1) as in part a), the pAg is:<br />pAg = -log(0.0980) = 0.9<br /><br />Therefore, the correct answer is:<br />a) pAg = 1<br />b) pAg = 1.3<br />c) pAg = 0.9