In titration of 10.00 mL of an unknown weak acid (HX) with astrong base that has
ID: 677399 • Letter: I
Question
In titration of 10.00 mL of an unknown weak acid (HX) with astrong base that has a molarity of 0.09994 M, the equivalencevolume was found to be 21.20 mL. a) Calculate the concentration of HX b) If the initial pH of the solution was recorded as 2.92,what is the Ka and pKa of HX. In titration of 10.00 mL of an unknown weak acid (HX) with astrong base that has a molarity of 0.09994 M, the equivalencevolume was found to be 21.20 mL. a) Calculate the concentration of HX b) If the initial pH of the solution was recorded as 2.92,what is the Ka and pKa of HX.Explanation / Answer
a) Given data Volume of HX, V1=10.00 mL Molarity of HX, M1 = ? Volume of Base, V2 = 21.2 mL Molarity of Base, M2 = 0.09994 M Molarity of HX, M1 = M2 *V2 / V1 = (10.0 mL)(0.09994M) / (21.2 mL) = 0.04714 M b) Given data pH of the solution = 2.92 [H+] = 10-2.92 = 0.0012 M HX -----------> H+ + A- Initital(M) 0.04714 0 0 Change(M) -x +x +x Equilibrium(M) 0.04714-x x x Giventhat x = 0.0012 M [HX]eq = 0.04714 - 0.0012 M = 0.04594 M Ka =[H+] * [A-]/ [HX]eq = (0.0012M)(0.0012M) / (0.04594 M) = 3.13*10-5 pKa = - log(3.13*10-5) = 4.5 Initital(M) 0.04714 0 0 Change(M) -x +x +x Equilibrium(M) 0.04714-x x x Giventhat x = 0.0012 M [HX]eq = 0.04714 - 0.0012 M = 0.04594 M Ka =[H+] * [A-]/ [HX]eq = (0.0012M)(0.0012M) / (0.04594 M) = 3.13*10-5 pKa = - log(3.13*10-5) = 4.5Related Questions
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