Chem119-935am-quiz9-key

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Chemistry

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Dec 6, 2023

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Name:___________________________ Student ID:______________________________ Barondeau Chem 119 9:35 am class Quiz 9 1. You mix 55 mL of 1.50 M silver nitrate (AgNO 3 ) with 35 mL of 1.47 M sodium chloride (NaCl). Determine the mass (with 2 significant figures) of precipitated silver chloride (AgCl; 143.32 g/mol) that would form (2 points). AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) Ag + (aq) + NO 3 - (aq) + Na + (aq) + Cl - (aq) AgCl(s) + Na + (aq) + NO 3 - (aq) Remove spectator ions Ag + (aq) + Cl - (aq) AgCl(s) Limiting reagent problem (1.50 moles AgNO 3 / L)(0.055 L) = 0.0825 moles AgNO 3 or Ag + (1.47 moles NaCl / L)(0.035 L) = 0.05145 moles NaCl or Cl - Cl - is limiting reagent (0.05145 moles Cl - )(1 mol AgCl / 1 mol Cl - )(143.32 g AgCl / mol) = 7.3738 or 7.4 g 2. You have 51.5 mL of a 2.80 M solution of Na 2 CrO 4 ( aq ). You also have 112 mL of a 2.50 M solution of AgNO 3 ( aq ). Calculate the concentration of Na + after the two solutions are mixed. Report the answer with 3 significant figures (2 points) Determine number of moles Na+ and then divide by total volume. (2.80 moles Na 2 CrO 4 / L)(0.0515 L)(2 Na + /Na 2 CrO 4 ) = 0.2884 moles Na + Total volume = 51.5 mL + 112 mL = 163.5 mL Concentration = (0.2884 moles / 0.1635 L) = 1.76 M Na + 3. The hydronium ion concentration in a solution was measured to be 4.8 10 -6 M. a. What is the pH of this solution? Report the answer with 2 significant figures (1 point). pH = -log [H + ] pH = -log (4.8 10 -6 M) = 4.7 b. Is this an acidic or basic solution (1 point)? At room temperature: pH below 7 is acidic. pH above 7 is basic. Acidic 7.4 g 1.76 M 5.3
4. Write a net ionic equation for the reaction of the weak acid HF with the strong base KOH (2 points) HF(aq) + KOH(aq) KF(aq) + H 2 O( l ) HF(aq) + K + (aq) + OH - (aq) K + (aq) + F - (aq) + H 2 O( l ) Remove spectator ions HF(aq) + OH - (aq) F - (aq) + H 2 O( l ) 5. Calculate the mass of NaOH (39.997 g/mol) needed to completely neutralize 58.3 mL of a 0.240 M solution of the monoprotic acid HNO 3 . Report the answer with 3 significant figures (2 points). NaOH(aq) + HNO 3 (aq) NaNO 3 (aq) + H 2 O( l ) (0.240 moles HNO 3 / L)(0.0583 L) = 0.013992 moles HNO 3 (0.013992 moles HNO 3 )(1 mol NaOH / 1 mol HNO 3 )(39.997 g NaOH / mol) = 0.559638 g HF(aq) + OH - (aq) F - (aq) + H 2 O( l ) 0.560 g
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