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Determine the percent ionization of a 0.125MHCN solution (K a
Determine the percent ionization of a 0.125MHCN solution (K a
=4.9×10 −10
), Express your answer using two significant figures.
Final answer:Percent ionizationof a 0.125M HCN solution are calculated first by setting up the ionization reaction and equilibrium expression, then solving for the concentration of the ionized HCN. By plugging these numbers into the percent ionization formula, it results to0.007% ionization.Explanation:To calculate thepercent ionizationwe first need to establish the chemical reaction, which would be:HCN <--> H+ + CN-. For this weak acid (HCN), we can assume that the initial concentration of ions is negligible, so the initial concentration of HCN is 0.125M. The ionization of HCN in water isgovernedby its acid ionization constant, Ka = 4.9 x 10^-10, which is expressed by the following equation: (H+)(CN-) / HCN = Ka.Since theconcentrationsof H+ and CN- are the same, let's call these x. The equation becomes x^2/0.125 = 4.9 x 10^-10. Solving this forx(which represents the concentration of ionized HCN) we get x = 8.78 x10^-6.The percent ionization iscalculatedas (concentration of ionized acid/initial concentration of acid) x 100%. Therefore, the percent ionization of the 0.125M HCN solution is(8.78 x 10^-6 / 0.125) x 100% = 0.007%Learn more aboutPercent Ionizationhere:brainly.com/question/32389094#SPJ11...