Consider a 0.10 M solution of a weak polyprotic acid ( H2A

Problem 116 EConsider a 0.10 M solution of a weak polyprotic acid with the possible values of and given here.a.b.c.Calculate the contributions to from each ionization step. At what point can thecontribution of the second step be neglected Step by step solutionStep 1 of 3(a)The reaction of polyprotic acid in aqueous solution is as follows:The concentration of is 0.10 M, equilibrium constant for first, and second ionizationstep, is and respectively.Consider the ICE table to calculate the change in concentration as follows:Calculate the equilibrium constant from equation (1) as follows:Substitute the values from ICE table.\nsince, equilibrium constant is smaller than initial concentration thus, neglects x from thedenominator in the above equation.ThusRearrange to get the value of x OrTherefore, from first ionization step concentration of isNow, consider the reaction between and as follows:Consider the ICE table to calculate the change in concentration by taking concentration of and asCalculate equilibrium constant from equation (1) as follows:Substitute the values from ICE table,\nSince, value of equilibrium constant is comparable to initial concentration thus, y cannot beneglected.Rearrange the above equation to get the value of y as follows:Solve the quadratic equation by quadratic formula as follows:Substitute, 1 for a, for b and for cThus =Concentration cannot be negative therefore, value of y is .Therefore, concentration of from second ionization step is .Now, calculate the ratio, R of concentration of from second, and firstionization step, \left[ H _{3} O \right]_{1}^{+}as follows:Substitute, for and for in aboveequation.Thus\nTherefore, the ratio is less than 0.05 or 5% therefore, second contribution can beneglected.