How would increasing the molarity of the Na OH solution from 0.01M to 0.1M affect the number of drops required?
Q. How would increasing the molarity of the Na OH solution from 0.01M to 0.1M affect the number of drops required for the observed color changes?
Asked by 123 - Sat Oct 30 16:54:48 2010 - Chemistry - 2 Answers - Comments

A. it would decrease the number of Na OH drops required for example: we have a 0.01M HCl solution. this reacts with Na OH in a 1:1 fashion. therefore, 0.01moles HCl is neutralized by 0.01moles Na OH. if we had 10ml of HCl, we would need 10ml Na OH now, take that 0.01M HCl solution, it still reacts with Na OH in a 1:1 fashion. if we had 10ml of 0.01M HCl this would be 0.1moles HCl in solution. if we neutralized this with 0.1M Na OH, we would need 1ml Na OH to neutralize the solution because 1ml 0.1M Na OH = 0.1moles Na OH, the amount of HCl needed to be neutralized.
Answered by Caroline Miller - Sat Oct 30 17:22:54 2010

Basic chemistry: how much Na OH required to change p H of a 1 liter p H 4 solution?
Q. How many milliliters of a 1 Molar Na OH solution would be required to change a 1 liter solution from p H 4 to p H 10? How many grams of Na OH would this translate to?
Asked by combatwombat - Fri Jan 16 21:50:00 2009 - Chemistry - 1 Answers - Comments

A. 1 liter of p H 4 solution is 10^-4 M in H+ = 0.1mmol of H+ p H 10 = 10^-4 M OH so adding 0.2 m L of 1 M Na OH will produce 0.2 mmoles of OH, neuralizing the 0.1 mmole of H+ and leaving 0.1 mmole of OH^-1 in 1000.2 m L 0.1 mmole of OH^-1 =10^-4 moles/1.002 liters = 0.0009998 M = 10^-4 M p OH = -log 10^-4 =4 p H =14-4 = 10
Answered by Merlin's Feline - Fri Jan 16 22:06:31 2009

How do you find the molarity of Na OH solution for a monopratic acid?
Q. the question goes as follows: 17.73 m L of an Na OH solution are needed to titrate 0.4532 g of a monoprotic acid (molecular mass = 122.1 g/mol) to end point. What is the molarity of the Na OH solution?
Asked by allchem'dout - Sun Nov 29 20:16:12 2009 - Chemistry - 1 Answers - Comments

An unknown amount of water is mixed with 350 m L of a 6 M solution of Na OH solution. A 75 m L sample of the resu?
Q. lting solution is titrated to neutrality with 52.5 m L of 6 M HCl. Calculate the concentration of the diluted Na OH solution. Answer in units of M. What volume of water was added to the 350 m L of Na OH solution? Assume volumes are additive. Answer in units of m L.
Asked by astrorocket - Wed Jan 7 21:53:05 2009 - Chemistry - 1 Answers - Comments

A. Moles HCl = 6 M x 0.0525 L= 0.315 = moles Na OH in 75 m L Initial concentration = 0.315 x 350/ 75 = 1.47 M
Answered by Dr.A - Thu Jan 8 12:58:07 2009

What mass of oxalic acid is required to react with 40m L of 0.100M Na OH solution?
Q. What mass of oxalic acid is required to react with 40m L of 0.100M Na OH solution? The oxalic acid is completely neutralized to oxalate ion.
Asked by ashpinkpuppy21 - Sat Oct 22 19:08:36 2011 - Chemistry - 1 Answers - Comments

A. Since oxalic acid (modern IUPAC name ethanedioic acid) is a dibasic acid, 1 mole oxalic acid reacts with 2 moles Na OH. Moles of Na OH = vol in L x molarity = 0.040 x 0.1 moles So will react with half that number of moles of oxalic acid. Mass of oxalic acid = moles x molar mass Molar mass of oxalic acid =90g Answer = 1/2x 0.040 x 0.1 x 90g
Answered by Colin - Wed Oct 26 17:26:24 2011

How do I determine the concentration of a Na OH solution using calorimetry?
Q. I've got a solution of Na OH approximately 2 mol dm-3 in concentration. How do I use a calorimeter to accurately determine it's concentration and what calculations do i need to do?
Asked by The Big Willi - Fri May 9 07:24:39 2008 - Chemistry - 1 Answers - Comments

A. neutralizing a strong base like Na OH with a strong acid like HCl releases heat equal to the heat released in: H+ (d Hf = 0) & OH- (d Hf = -230k J) --> H2O (d Hf = -285.83k J) d H reaction = d Hf's prod -reactants = -285.83 - (-230.0) = -55.8 kilojoules per mole of Na OH neutralized --- you take say 25 mls of your Na OH solution @ 2 mol/ litre which has about 0.005 moles... and you add an excess moles of HCl say 0.006 moles ... by adding 60 mls of 0.1 molar HCl measure the increase in temperature --- weight the amount of solution... determine the heat gained by the water: d H = mass of solution (4.184 J/gram-C)( temp change) change these Joules into kilojoules this amount of kilo Joules = that lost by the reaction divide those KJ by… [cont.]
Answered by Steve O - Fri May 9 13:57:02 2008

what volume of 3M stock Na OH solution you need to dilute ?
Q. what volume of 3M stock Na OH solution you need to dilute to 1 L to obtain an approx. 0.1 M solution ?
Asked by Mike - Tue Mar 2 22:07:30 2010 - Chemistry - 1 Answers - Comments

Caculate the total number of m L of the same Na OH solution required to completely neutralize the solution?
Q. 20.2m L of Na OH are required to reach the first endpoint in the titration of a solution of a diprotic acid H2A.
Asked by PyroRei - Sun Jun 29 22:13:33 2008 - Chemistry - 1 Answers - Comments

A. diprotic acid means you will need twice the number of moles of your hydroxide concentration. Please update your question with the exact words from wherever you got that, it is missing some key points that you need to solve the problem
Answered by Hyphy99 - Sun Jun 29 22:20:29 2008

What is the concentration of an Na OH solution?
Q. 50 milliliters of Na OH in solution at an unknown concentration is prepared. An indicator is then added. It took 23 milliliters of 0.2M H2SO4 solution to change activate the color change to indicate that all the Na OH had reacted. What is the concentration of Na OH solution? I'm having a lot of trouble remembering how to do this question. Any help would be appreciated!!
Asked by Ethan - Tue Apr 26 22:05:24 2011 - Chemistry - 1 Answers - Comments

How do you calculate the molarity of an aqueous Na OH solution?
Q. if 30.10 m L of .500 M H2SO4 is required to reach the end point when titrated against 10.00m L of the Na OH solution? The balanced chemical equation for this neutralization is- H2SO4 + 2 Na OH ---> Na2SO4 + 2 H2O Thank you for your help.
Asked by T-Rex84 - Thu May 31 05:27:10 2007 - Chemistry - 4 Answers - Comments

A. the mole of H2SO4 = 15.05mmoles thus the mole of Na OH should be 30.10 mmoles the molarity of Na OH = 30.10 : 10 = 3.01 M
Answered by Papilio paris - Thu May 31 05:35:02 2007

Na OH solution were to be standardized against pure solid primary standard grade KHP?
Q. If 0.4538g of KHP requires 44.12m L of the Na OH to reach a phthalein endpoint, what is the molarity of the Na OH solution?
Asked by Zer0 - Mon Jun 30 16:20:55 2008 - Chemistry - 1 Answers - Comments

A. 44.12 can go to grams since 1g in 1 ml...44.12 g * 1mol/ 39.9971g= 1.10308 mol...i dunno if this helps, but its a start! hehe
Answered by karmelcoloredprincess47 - Mon Jun 30 16:40:30 2008

What is the molar concentration of the Na OH solution?
Q. A 0.405 g sample of KHP is dissolved in 50 ml of water. The sample is then titrated with 15.50 ml of Na OH solution. What is the molar concentration of the Na OH solution?
Asked by TennisGirl - Thu Oct 27 20:54:46 2011 - Chemistry - 1 Answers - Comments

A. KHP and Na OH react in equimolar amounts. (0.405 g KHP) / (204.22 g KHP/mol) / (0.01550 L) = 0.128 M Na OH The 50 ml of water is not significant.
Answered by Roger the Mole - Thu Oct 27 21:46:41 2011

How many ml of the Na OH solution was required for the titration?
Q. 5.43g of Al(OH)3 to a beaker containing 250.0ml of a 1.00M HCl solution stirred it, and observed that the solid dissolved completely. A drop of indicator showed that solution was still acidic. then the solution was titrated with a 2.00M Na OH(aq). i really dont understand can someone help
Asked by missy2cute - Sun Nov 8 13:03:38 2009 - Chemistry - 1 Answers - Comments

A. The Al(OH)3 reacted with the HCl according to the following balanced equation: Al(OH)3 + 3 HCl = Al Cl3 + 3 H2O 1mol Al(OH)3 reacts with 3mol HCl Molar mass Al(OH)3 = 78.0037 g/mol 5.53g = 5.43 / 78.0037 = 0.0692mol Al(OH)3 This will react with: 0.0692*3 = 0.2077mol HCl In the 250ml 1.00M h Cl you have : 250/1000*1.00 = 0.25mol HCl 0.2077mol will be consumed in reacting with the Al(OH)3, so you have: 0.2500-0.2077 = 0.0423 mol HCl unreacted This will react with the Na OH Equation: Na OH + HCl Na Cl + H2O Na OH reacts with HCl in 1:1 molar ratio Therefore you require 0.0423 mol Na OH 1000ml Na OH 2.00M contains: 2.00mol Na OH Volume that contains 0.0423mol : 1000/2.00*0.0423 = 21.15ml 2.00M Na OH solution required.
Answered by Trevor H - Sun Nov 8 14:31:10 2009

What is the concentration of a Na OH solution if 32.47 m L of it are required to neutralize 1.27 g of KHC8H4O4?
Q. What is the concentration of a Na OH solution if 32.47 m L of it are required to neutralize 1.27 g of potassium hydrogen phthalate?
Asked by Prob - Sun Oct 10 10:20:45 2010 - Chemistry - 1 Answers - Comments

A. KHC8H4O4 + Na OH = KNa C8H4O4 + H2O 1 mole of phthalate reacts with 1 mole of hydroxide Molar mass of KHC8H4O4 is 204.2220 g/mol so moles of the standard = 1.27 / 204.222 = 0.006219 moles moles of Na OH at the eq point = 0.006219 these are in 32.47 mls so moles per liter = 0.006219 x 1000 / 32.47 = 0.192M
Answered by bob_the_chemist - Sun Oct 10 10:30:21 2010

Calculate the concentration of a NAOH solution given that 32.50 m L of it reacted with a solution of KHP which?
Q. Calculate the concentration of a NAOH solution given that 32.50 m L of it reacted with a solution of KHP which contained 0.375 g KHP.
Asked by Jane - Mon Nov 29 05:50:02 2010 - Chemistry - 1 Answers - Comments

A. moles KHP = 0.375 g / 204.227 g/mol=0.00184 = moles Na OH Molarity of Na OH = 0.00184 / 0.03250 L=0.0565 M
Answered by Dr.A - Mon Nov 29 11:24:34 2010

How many m L of 0.100M Na OH solution would be required to titrate a 0.145g sample of acetic acid to evuivalence?
Q. How many m L of 0.1000M Na OH solution would be required to titrate a 0.145g sample of acetic aci to equivalence point?
Asked by Michelle - Mon Nov 8 11:18:18 2010 - Chemistry - 2 Answers - Comments

A. moles acetic acid = 0.145 g/ 60.054 g/mol=0.00241 moles Na OH required = 0.00241 Volume Na OH = 0.00241/ 0.100 M = 0.0241 L => 24.1 m L
Answered by Dr.A - Mon Nov 8 12:01:43 2010

What would be the procedure for making 600m L of 0.5M Na OH solution if 3M of Na OH has already been provided?
Q. Also, How would I estimate how many m Ls of a 0.5 M Na OH solution will be required to titrate 1.2 g of potassium hydrogen phthalate to the phenolphthalein endpoint?
Asked by Billy - Thu Nov 12 14:43:38 2009 - Chemistry - 1 Answers - Comments

A. You need to dilute the 3M of Na OH down to 0.5M within a 600m L constraint. A simple formula is used C1V1=C2V2 C1=Concentration of provided (3M Na OH) V1=volume/amount of provided (unknown) C2=Concentration of desired (0.5M Na OH) V2=Volume/amount of desired (600m L) (Make sure all your units are the same...moles to moles and m L to m L) Then cross multiply to get your answer... (3M Na OH)(???m L 3MNa OH) = (0.5M Na OH)(600m L Na OH) You should need 100m L of 3M Na OH and 500m L of dilutent(probably water) to dilute a 3M Na OH solution down to 0.5M Na OH. If that doesn't help, try this dilution calculator...
Answered by Jinkies - Thu Nov 12 15:00:21 2009

How would you use 80g of sodium hydroxide (Na OH) to prepare a 1 m solution?
Q. Can someone help me make a step-by-step list for this problem? FACT: - one mole of Na OH has a molar mass of 40.0 g so 40.0 g of Na OH dissolved in 1 kg of water results in a one-molal Na OH solution.
Asked by csapxo - Thu Apr 15 22:51:04 2010 - Chemistry - 1 Answers - Comments

A. since approximately 40 g of Na OH = 1mole, 80 g = 2 moles. The formula for molarity is # of moles of solute/1 L of solvent. So to make a 1M solution of Na OH you must mix it with 80 g (2 moles) with 2 Liters of Water. 2 moles/ 2L = 1M To actually make the solution, first get a container that can hold at least 2L. Then pour in exactly 2L of water and make a precise mark at the bottom of the meniscus. Empty the water and dry. Next, place the 80g of Na OH in the same container. Finally add water to the container until the bottom of the meniscus reaches the mark you made earlier. Then mix. hope this helps
Answered by Andrew - Thu Apr 15 23:10:06 2010

What is the concentration of a Na OH solution?
Q. Potassium acid phthalate, KHC8H4O4, is a crystalline solid that is available in a state of high purity, making it an excellent choice as a standard acid. What is the concentration of a Na OH solution, if it takes 54.00 m L to neutralize a 0.0319 g sample of the acid? For the neutralization reaction, KHC8H4O4(aq) --> K+(aq) + H+(aq) + C8H4O42-(aq)
Asked by jack - Sat Oct 16 17:20:27 2010 - Chemistry - 1 Answers - Comments

A. The molar mass of potassium acid phthalate is 39 + (8 x 12) + (5 x 1) + (4 x 16) = 204 g.mol The number of moles of potassium acid phthalate is 0.0319 g/204 g/mol = 1.56 x 10^-4 mol The number of moles of H+ generated by the reaction shown is the same as the number of moles of the starting acid phthalate, 1.56 x 10^-4 mol. Neutralization would require the same number of moles of sodium hydroxide. The number of moles is related to the concentration (which you want to find) and the volume through n = c.v where n = number of moles. Rearranging, c = n/v c(Na OH) = 1.56 x 10^-4 mol/0.054 L = 2.89 x 10^-3 mol/L or M. The concentration of Na OH is 2.89 x 10^-3 M or 0.29 m M.
Answered by Andrew - Wed Oct 20 03:35:28 2010

What was the molarity of the Na OH solution?
Q. A volume of 12.43 ml of a freshly prepared solution of Na OH was required to titrate 15.00 ml of 0.316 M H2SO4 solution. show steps please.
Asked by katie - Mon Oct 12 21:49:13 2009 - Chemistry - 1 Answers - Comments