Question

Question 4 2 pts What is the molarity of the OH in a solution obtained by mixing 125 mL 2.35 M Ca(OH)2 and 355 mL 5.52 M Ba(OH)2 5.31M 2.65 M 10.62 1.33

Answer 1

given that Volume Molarity OE volume Molarity of CaCOH)2 = 125ML CaloH)2 = 2.35M BalOH)2 = 355 ML Balot) a = 5.52 M ولی دلم we Know that Moles of solute molarity of solution = Volume of solution (in L) Moles of solute = molarity of Solution X volume of solution (2) molarity of Calota = 2.35 M=2.35 mal Volume of Calolz=125ML since, Cime L = 0.ool 0.001L). Volume OF caloH)2 =12580.00lL - 0.125L 22:35 mol ولاه لم Moles of caloH)2 = molarity of caloH? X Volume of CalOH)2 Moles of caloH)₂ = 2.35 mol xo.12.5X 0.294 mol moles Са (он) — 0.294 mol Since, 1.0 mol of CaloH) contain 2 mol of OH ion 0.294 mol of Ca(OH)2 Contain 2x0.294 Mol OH = 0.588 mal ot con Hence, moles OF ot ion in 0.294 mor CaCOH)2 = 0.588 Mol

Nowi Volume Volume BalOH)2 = 355x0.ooll Ba(OH)2 = 0.355 L BalOH)2 = 5.52 mol/l OF molarity dow, Moles BaloH)2 = Molarity volume of BALOH)2 = 5.52 mol xo.355€ t Moles of Balote = 1.96 mol since, 1 mol of 1.96 mol of BaCOH)2 contain a mol of oth ion BalOH)2 contain ax1.96 mol of OH-ion = 3.92 mol of OH ion Hence, moles Ba(OH)2 of OH con in 1.96 mol -13.92 mol of OH ot in Total moles in solution - Sum of moles OF CaloH) and BalGH) 0.588 mol + 3.92 mol = 4.508 mal Total Modes of oh in solution = 4.508 mod) Volume of solution - Volume of CalOH)2 + volume OF Ba(OH)2 = 0.1254+0.355L. Volume of Solution = {0.48L ولي ولم molarity of 얘 - Moles of oH (Total) Volume of solution (int)

4. Бо мо я 6. Че 9. Чечь 4 • — 9. Чом Hence, Molarity 19. Дом solution Answer.