Interpretation:
The number of moles of lead
Concept introduction:
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Introductory Chemistry: An Active Learning Approach
- Potassium acid phthalate, KNaC8H4O4, or KHP, is used in many laboratories, including general chemistry laboratories, to standardize solutions of base. KHP is one of only a few stable solid acids that can be dried by warming and weighed. A 0.3420-g sample of KNaC8H4O4 reacts with 35.73 mL of a NaOH solution in a titration. What is the molar concentration of the NaOH? KNaC8H4O4(aq)+NaOH(aq)KNaC8H4O4(aq)+H2O(aq)arrow_forwardRelative solubilities of salts in liquid ammonia can differsignificantly from those in water. Thus, silver bromide issoluble in ammonia, but barium bromide is not (thereverse of the situation in water). Write a balanced equation for the reaction of anammonia solution of barium nitrate with an ammoniasolution of silver bromide. Silver nitrate is soluble inliquid ammonia. What volume of a 0.50 M solution of silver bromidewill react completely with 0.215 L of a 0.076 M solutionof barium nitrate in ammonia? What mass of barium bromide will precipitate fromthe reaction in part (b)?arrow_forwardLaws passed in some states define a drunk driver as one who drives with a blood alcohol level of 0.10% by mass or higher. The level of alcohol can be determined by titrating blood plasma with potassium dichromate according to the following equation 16H+(aq)+Cr2O72(aq)+C2H5OH(aq)4Cr3+(aq)+2CO2(g)+11H2O Assuming that the only substance that reacts with dichromate in blood plasma is alcohol, is a person legally drunk if 38.94 mL of 0.0723 M potassium dichromate is required to titrate a 50.0-g sample of blood plasma?arrow_forward
- Lead(II) nitrate is added to four separate beakers that contain the following: aker 1 (sodium chloride) eaker 2 (sodium hydroxide) eaker 3 (sodium phosphate) eaker 4 (sodium sulfate) ter the addition of the lead(II) nitrate solution to each beaker, in which beaker(s) will a precipitate form? Use the general solubility rules given in Table 7.1 to guide you.arrow_forwardA student was given a 1.6240-g sample of a mixture of sodium nitrate and sodium chloride and was asked to find the percentage of each compound in the mixture. She dissolved the sample and added a solution that contained an excess of silver nitrate. The silver ion precipitated all of the chloride ion in the mixture as silver chloride. It was filtered, dried, and weighed. Its mass was 2.056g. What was the percentage of each compound in the mixture?arrow_forwardDescribe the use of gravimetric analysis to determine the percentage of chlorine in a water-soluble unknown solid.arrow_forward
- Twenty-five milliliters of a solution (d=1.107g/mL)containing 15.25% by mass of sulfuric acid is added to 50.0 mL of 2.45 M barium chloride. (a) What is the expected precipitate? (b) How many grams of precipitate are obtained? (c) What is the chloride concentration after precipitation is complete?arrow_forwardHow many grams of precipitate are formed if 175 mL of a 0.750 M aluminum sulfate solution and 375 mL of a 1.15 M sodium hydroxide solution are mixed together?arrow_forwardDecide whether a precipitate will form when the following solutions are mixed. If a precipitate forms, write a net ionic equation for the reaction. (a) potassium nitrate and magnesium sulfate (b) silver nitrate and potassium carbonate (c) ammonium carbonate and cobalt(lll) chloride (d) sodium phosphate and barium hydroxide (e) barium nitrate and potassium hydroxidearrow_forward
- On the basis of the general solubility rules given in Table 7.1, predict which of the following substances are not likely to be soluble in water. Indicate which specific rule(s) led to your conclusion. :math>PbSe. BaCO3 :math>Mg(OH)2f. AlPO4 :math>Na2SO4g. PbCl2 :math>(NH4)2Sh. CaSO4arrow_forwardWhat mass of silver chloride can be prepared by the reaction of 100.0 mL of 0.20 M silver nitrate with 100.0 mL of 0.15 M calcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete.arrow_forwardA student added 50.0 mL of an NaOH solution to 100.0 mL of 0.400 M HCl. The solution was then treated with an excess of aqueous chromium(III) nitrate, resulting in formation of 2.06 g of precipitate. Determine the concentration of the NaOH solution.arrow_forward
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