It required 66.66 mL of 2.000 M HBRO3 to titrate (neutralize) 50.00 mL of an Al(OH)3 solution. What is the molarity, M, of the aluminum hydroxide solution ?З HBГОз + Al(ОН)з3 H20 + Al(BrO3)3Answer a few questions before giving the final answer in the last problem.There is enough information to calculate theA. moles of Al(OH)3B. moles of HBRO3

Molarity (M) = number of moles of solute (n) / Volume of solution in litre at neutralization point moles of acid = moles of base Ma x Vana = Mb x Vbnb where Ma = molarity of acid = 2 M Mb = molarity of base = ? Va = Volume of acid = 66.66 mL Vb = Volume of base = 50 mL na = moles of acid = 3 mol nb = moles of base = 1 mol 2 M x 66.66 mL3 mol = Mb x 50 mL1 mol 44.44 M = 50 x Mb Mb = 44.44 50 M = 0.88 M (A) moles of Al(OH)3 = M base x Vbase = 0.88 M x ( 50 mL x 1 L/ 1000 mL) = 0.044 mol (B) moles of HBrO3 = M acid x Vacid = 2 M x ( 66.66 mL x 1L / 1000 mL ) = 0.132 mol

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It required 66.66 mL of 2.000 M HBrOg to titrate (neutralize) 50.00 mL of an Al(OH)3 solution. What is the molarity, M, of the aluminum hydroxide solution ? 3 HBRO3 + Al(OH)3 → 3 H20 + Al(BrO3)3 Answer a few questions before giving the final answer in the last problem. There is enough information to calculate the

It required 66.66 mL of 2.000 M HBrOg to titrate (neutralize) 50.00 mL of an Al(OH)3 solution. What is the molarity, M, of the aluminum hydroxide solution ? 3 HBRO3 + Al(OH)3 → 3 H20 + Al(BrO3)3 Answer a few questions before giving the final answer in the last problem. There is enough information to calculate the

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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