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Chemistry questions on finding OH, pH, and pKa?

  • I have worked numerous questions like these and for some reason I am stumped on these!! Calculate [OH-] for a solution formed by adding 6.00 mL of 0.180 M KOH to 10.0 mL of 8.3×10−2 M Ca(OH)2. Calculate [OH-] for a solution formed by adding 6.00 mL of 0.180 M KOH to 10.0 mL of 8.3×10−2 M Ca(OH)2. This is also confusing me - I think I had used the wrong equation: A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research technician at Pharma-pill, you are told that X-281 is a monoprotic weak acid, but because of security concerns, the actual chemical formula must remain top secret. The company is interested in the drug's Ka value because only the dissociated form of the chemical is active in preventing cholesterol buildup in arteries. To find the pKa of X-281, you prepare a 0.088 M test solution of X-281. The pH of the solution is determined to be 2.40. What is the pKa of X-281? I came up with 4.94 and it was wrong! Any help greatly appreciated!!!

  • Answer:

    Q1. Find the total number of moles of OH- and divide by the Liters of solution. moles KOH = M KOH x L KOH = (0.180)(0.00600) = 0.00108 moles KOH Since there is 1 OH- in KOH, then moles [OH-] from KOH = 0.00108 moles Ca(OH)2 = M Ca(OH)2 x L Ca(OH)2 = (0.083)(0.0100) = 0.00083 moles Ca(OH)2 Bedcause there are 2 OH- in Ca(OH)2, then moles OH- from Ca(OH)2 x 2 x 0.00083 = 0.00166. Total moles OH- = 0.00108 + 0.00166 = 0.00274 [OH-] = 0.00274 moles OH- / 0.01600 L solution) = 0.171 M ======================================… Q2. X-281 (which I'll call HX) will react with water to form small but equal amounts of H3O+ and X-. Molarity . . . . . . . .HX + H2O <==> H3O+ + X- Initial . . . . . . . . . 0.088 . . . . . . . . . . . .0 . . . . .0 Change . . . . . . . . .-x . . . . . . . . . . . . . x . . . . .x At Equilibrium . . .0.088-x . . . . . . . . . . x . . . . . x So if the pH = 2.40, then [H3O+] = 10^-pH = 10^-2.40 = 4.0 x 10^-3 = 0.0040 = x. Ka = [H3O+][X-] / [HX] = (x)(x) / (0.088-x) = (0.0040)2 / (0.088 - 0.0040) = 1.9 x 10^-4 pKa = -log Ka = -log (1.9 x 10^-4) = 3.72

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Ca(OH)2 will dissociate Ca(OH)2 ↔ Ca2+ + 2OH- If a large concentration of OH- ions is introduced to the solution, the common ion effect must be considered. This extra concentration of OH- ions causes the above equilibrium to shift to the left. The concentration of OH- ions originating from the Ca(OH)2 dissociation is reduced. The two concentrations are not additive. It is necessary to calculate the solubility of Ca(OH)2 in a solution containing excess OH- ions First calculate the concentration of KOH from the 6.00mL of 0.180M KOH in 16mL solution M1V1 = M2V2 M1 * 16 = 0.180*6.0 M1 = 0.0675M KOH solution Now calculate the solubility of Ca(OH)2 in this solution: Ksp for Ca(OH)2 = 7.9*10^-6 Ksp = [Ca2+] [OH-]² We know that the [OH-] from the KOH = 0.0675M and that the [OH-] from the Ca(OH)2 is very small and can be ignored in this equation 7.9*10^-6 = [Ca2+] * (0.0675)² 7.9*10^-6 = [Ca2+] * 4.56*10^-3 [Ca2+] = (7.9*10^-6) / ( 4.56*10^-3) [Ca2+] = 1.73*10^-3 If the Ca2+ = 1.73*10^-3 , then OH- from this dissolved Ca(OH)2 = 3.46*10^-3 Total concentration of OH- in solution = 0.0675M from the KOH dissociation, and 3.46*10^-3 M from the dissociation of the Ca(OH)2. Add these together: 6.75*10^-2 + 3.46*10^-3 = 7.1*10^-2M The Ca(OH)2 has made only a small contribution to the [OH-] because of the common ion effect. First determine the [H+] pH = 2.40 [H+] = 10^-2.4 [H+] = 3.98*10^-3 Call the acid HA Calculate Ka Ka = [H+] [A-] / [HA] Because [H+] = [A-] , and [H+] is very small compared to [HA], we can substitute: Ka = [H+]²/[HA] Ka = (3.98*10^-3) ² / 0.088 Ka = 1.58*10^-5/ 0.088 Ka = 1.8*10^-4 pKa = - log Ka pKa = -log (1.8*10^-4) pKa = 3.74

Trevor H

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