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Lindsay

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molar concentration problems

« on: October 21, 2008, 01:36:53 PM » Hello,I need some help with various problems. Please make the answers are correct1. A 4 g mass of NaOH is dissolved in 5mL of watera. what is the approximate molar concentration of the NaOH?My answer..4 g X 1 mol NaOH/40.00 g NaOH = .1 mol NaOHM=m/V .1 mol NaOH/.005L = 20 Mb. a 4 mL aliquot of this solution is diluted to 500 mL of solution. What is the approximate molar concentration of NaOH in the diluted solution?m=M X V20M X .004 = .08 m NaOHM= .08m/.5L = .16 Mc. Calculate the mass of KHC8H4O4 that reacts with 15 mL of the NaOH solution KHC8H4O4 + NaOH ----- H2O + NaKC8H4O415 mL/1000 X .16 M NaOH/ 1L X 1 mol KHC8H4O4/1 mol NaOH X 204.44 g KHC8H4O4/ 1 mol KHC8H4O4 = 0.49 g KHC8H4O4Thanks,Lindsay Logged

Astrokel

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Re: molar concentration problems

« Reply #1 on: October 21, 2008, 01:57:13 PM » Correct! well done just a side note:Quote

m=M X V20M X .004 = .08 m NaOH

i don't know if they will penalized you for using m instead of n (m = M X V) as we usually use m for mass and n stands for moles. But your SI unit for moles is wrong it is not m, it is in mol Logged No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Lindsay

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Re: molar concentration problems

« Reply #2 on: October 21, 2008, 02:51:46 PM » Thanks!I have two more problems.Here they are1. (A) 0.4040 g sample of potassium hydrogen phthalate, KHC8H4O4 ( molar mass = 204.44) is dissolved with 50 mL of deionized water in a 125 mL flask. The sample is titrated to the phenolphthalein endpoint with 14.71 mL of a sodium hydroxide solution. What is the molar concentration of the NaOH solution?My answer is 0.4040 g KHC8H4O4 X 1 mol KHC8H4O4/204.44 g KHC8H4O4 X 1 mol NaOH/ 1 mol KHC8H4O4 = 0.001976 mole of NaOHM= Moles of Solute/L of solutionM = 0.001976/ 0.01471 L = 0.1343 M of NaOH(B) A 25.000 mL aliquot of a nitric acid solution of unknown concentration is pipetted into a 125 mL flask and 2 drops of phenolphthalein are added. The above NaOH solution (the titrant) is used to titrate the nitric acid solution (the analyte). If 18.92 mL of the titrant is dispensed from a buret in causing a color change of the phenolphthalein, what is the molar concentration of the nitric acid solution?I'm not sure on this one. The 2 drops are throwing me off. you have a reaction ofHNO3 + NaOH----H2O + NaNO3 so18.92/1000 X 0.1343/1 L Solution X 1 mol HNO3/ 1 mol NaOH = 0.002541 mole of HNO3Then, I guess, you need to find the molar conc of the acidM = moles of solute/ Liters of SolutionM = 0.002541/ 0.25 L = 0.01016I'm not sure about this two problems. Please advice.Thanks,Lindsay Logged

Astrokel

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Re: molar concentration problems

« Reply #3 on: October 21, 2008, 03:00:27 PM » Both are correct!  The 2 drops added are just indicator and it doesn't affect your end point. Logged No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

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