Please help i dont know what to do! In a certain experiment, 28.01 mL of .2500 M
ID: 691127 • Letter: P
Question
Please help i dont know what to do! In a certain experiment, 28.01 mL of .2500 M Nitric Acid and53.0 mL of 0.3204 M Potassium hydroxide are mixed. A)calculate the amount of water formed in the resultingreaction. B) what is the concentration of Hydrogen Ions or HydroxideIons in excess after the reaction is complete? Please help i dont know what to do! In a certain experiment, 28.01 mL of .2500 M Nitric Acid and53.0 mL of 0.3204 M Potassium hydroxide are mixed. A)calculate the amount of water formed in the resultingreaction. B) what is the concentration of Hydrogen Ions or HydroxideIons in excess after the reaction is complete?Explanation / Answer
A) The equation for the reaction: HNO3 + KOH -----> KNO3 + H2O . The equation shows one mole of acid and base reacting to give onemole of water. . Moles of nitric acid = molarity * volume (L) = 0.2500 M * 0.02801 L = 0.007 moles . Moles of potassium hydroxide = 0.3204 M * 0.053 L = 0.017 moles . So nitric acid is the limiting reagent in the reaction 0.007 moles of nitric acid react with 0.007 moles of KOH to give0.007 moles of water. Mass of water = moles * molar mass = 0.007 moles * 18.02g/mol = 0.126 g . B) Moles of excess KOH = moles of hydroxide ions = 0.017 - 0.007 =0.01 moles Total volume of solution = 28.01 mL + 53.0 mL = 81.01 mL Since density of water ˜ 1 g/mL, volume of water produced =0.126 mL So total volume = 81.01 + 0.126 = 81.136 mL . Concentration of hydroxide ions = moles * 1000 / volume (mL) = 0.01 moles * 1000 / 81.136 mL = 0.123 M
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