How many mililiters is of Na OH was required to neutralize the acid?
Q. 10 m L of a 1 M HCl solution was added to a flask and titrated against a solution of 2.1 M Na OH.
Asked by symia87 - Tue Nov 20 12:41:46 2007 - Chemistry - 2 Answers - Comments

A. Moles HCl = 10 x 1 / 1000 = 0.01 Moles Na OH = moles HCl M = mol / L L = mol / M = 0.01 / 2.1 = 0.00476 L => 4.76 m L
Answered by Dr.A - Tue Nov 20 14:35:09 2007

Why is it important to have the temperatures of both HCl and Na OH solutions close together?
Q. So going through my stuff from this past school year i found a question that I had answered wrong but i can't figure out what i was missing in the answer... In a Laboratory experiment using solutions of HCl and Na OH (made with 90% water) why is it important to have the temperatures of both HCl and Na OH solutions close together?
Asked by MayMay - Tue Jul 19 16:26:01 2011 - Chemistry - 1 Answers - Comments

A. Because if you don t, how are you going to decide on what the initial temperature was?
Answered by Visman Mickus - Tue Jul 19 16:29:27 2011

How many moles of Na OH are required to titrate .0353 moles of khp to the equivlaence point?
Q. Just wondering if you could help with this problem, How many moles of Na OH are required to titrate .0353 moles of khp to the equivlaence point? Also how many ml of 0.0964 M Na OH are required to titrate 0.0354 moles of KHP to the equivalence point? Thanks!
Asked by Jessie - Sun Apr 10 17:49:01 2011 - Chemistry - 1 Answers - Comments

A. Assuming the khp assay is high - 100% - 0.0353 moles of Na OH are required : The KHP:Na OH mole ratio is 1:1 That should enable you to answer the second part of your question.
Answered by Doc89891 - Sun Apr 10 17:54:28 2011

How would you find the concentration of a diprotic acid that was titratred with Na OH?
Q. Use the method you determined to find the concentration of a diprotic acid when 24.55 m L of the acid requires 32.57 m L of 0.1059 M Na OH.
Asked by Patrick Ryan - Wed Apr 27 15:54:49 2011 - Chemistry - 1 Answers - Comments

A. Well diprotic acids react 1:1 with Na OH to form a salt and water. If you find the amount of moles of Na OH used in the titration, it will be the same as the amount of moles of H+. moles of OH- = (.0357 L) * (.1059 M) = .00345 moles of OH. Again since it required .00345 moles of OH to react with an acid, the acid had .00345 moles of H+ to contribute to the reaction. Molarity = moles/L M= .00345 moles H+/.02455 L = .141 M Hope this helps!
Answered by Motown Born - Wed Apr 27 16:00:37 2011

What is the balanced equation of methyl salicylate and Na Oh? How do I rationalize this question?
Q. Using the balanced equation for the reaction of methyl salicylate with sodium hydroxide, calculate the theoretical number of millimoles of deprotonated salicylic acid are formed from the reaction of 57 millimoles of methyl salicylate are reacted with 21 m L of 4.0 M Na OH. The answer is 28 mmoles but I don't understand why.
Asked by daye2sexy4u - Thu Sep 22 13:33:03 2011 - Chemistry - 1 Answers - Comments

A. The benzene ring contains an OH and a COOCH3 group Na OH replaces the CH3 which reacts with the OH grup to form methanol, so we end ujp up with soduim salicylate. Moles of Na OH = 0.021 x 4.0 = 0.084 moles. which is 84 millimoles is in excess. so the product is 57millimoles
Answered by John W - Thu Sep 22 14:22:14 2011

How many moles of Na OH are needed each day for each person in the spacecraft?
Q. As early as 1938, the use of Na OH was suggested as a means of removing CO2 from the cabin of a spacecraft according to the following reaction: Na OH + CO2 -> Na2CO3 + H2O. If the average human body discharges 925.0 g of CO2 per day, Question: How many moles of Na OH are needed each day for each person in the spacecraft?
Asked by Kara - Sat Feb 13 16:52:43 2010 - Chemistry - 1 Answers - 1 Comments

A. Step 1: Work out the moles of the subtance you know the mass of. In this case it is CO2. Work out its relative molecular mass using atomic masses from the periodic table: C = 12 O = 16 Mr = 12 + (2*16) = 44 Then use the equation Moles = Mass / Mr: Moles = 925 / 44 = 21.0227... Step 2: Use the balanced equation to work out the ratio of CO2 to Na OH. Neither of them have a balancing number so the ratio is simply 1:1 Step 3: Use the ratio to work out moles of Na OH. Because the ratio is 1:1, there will be the same number of moles of Na OH as CO2. Therefore, Moles of Na OH = 21.0227... = 21.02 (2dp)
Answered by FugueDude - Sat Feb 13 17:14:38 2010

What happens if you use Na OH to wash ester?
Q. Instead of sodium bicarbonate, what would happen if you wash ester product with Na OH? It would produce more alcohol but what does that ultimately do? Thanks for the help!
Asked by My avatar has a mullet! - Mon Jan 26 22:28:51 2009 - Chemistry - 2 Answers - Comments

A. It causes a reaction called saponification, which is the reverse reaction of the formation of the ester. The products would be the sodium salt of the acid and the alcohol that were used to form the ester. This process was used to make soap from lye and animal fat in pioneer days.
Answered by TheProf - Mon Jan 26 22:34:40 2009

How do you figure out emperical formula from a back titration involving HCl and Na OH?
Q. Say, for instance, you have a Group 1 metal hydroxide of a known mass, and you want to figure out its empirical formula by doing an experiment where you add a standard HCl solution to it, then titrate it with a standard Na OH solution. How would you figure it out?
Asked by Core Blimey - Fri Jun 18 23:18:46 2010 - Chemistry - 1 Answers - Comments

A. The empirical formula of group I metal hydroxide is always MOH. But if you want to determine the identity of your metal X... Weight an amount of this unknown MOH where M = group I metal. Dissolve in water. Add standard HCl. Back titrate excess HCl (unreacted) with standard Na OH. All reactions follows 1 mol base:1 mol acid ratio. Mole MOH = Mole HCl supplied - moles HCl unreacted (which is equal to mole Na OH used in back titration) mole of MOH = 1 mol OH = mole of M mass of OH = mole MOH x (17 g OH/ 1 mol OH) mass of metal = mass of sample - mass of OH Hence the molar mass of metal M = mass of metal/ mole of MOH Compare this value with the molar mass of metal in group I.
Answered by Flongkoy - Fri Jun 18 23:36:02 2010

What are the products created when Na OH and Cu SO4 react? At different concentrations?
Q. What is the difference in products when Na OH is either a low or a high concentration?
Asked by NEEDs MATH HELP!! - Tue Oct 16 18:52:09 2007 - Chemistry - 1 Answers - Comments

A. 2Na OH + Cu SO4 => Na2SO4 + Cu(OH)2
Answered by kitten - Tue Oct 16 18:58:12 2007

What is the molarity of a solution that contains 80.0 grams of Na OH dissolved in 500 m L of solution?
Q. What is the molarity of a solution that contains 80.0 grams of Na OH dissolved in 500 m L of solution?
Asked by princess_bear08 - Sat Apr 3 15:59:11 2010 - Chemistry - 1 Answers - Comments

A. Molarity is in M/L so first you have to determine how many moles are present. Na OH has a mol wt of 40g/mole so here you have 2 moles of Na OH. It is a 500m L solution, so you'll have to double that to get the target 1L. Here's how I did it: 40g/mol---> 2mol/500m L X 1000ml/L = 4mol/L so the molarity is 4 moles per liter. Hope that helped
Answered by garlanbc - Sat Apr 3 16:09:05 2010

When phenol is treated with CHCl3 and Na OH followed by acidification salicylaldehyde is obtained. Which of the?
Q. When phenol is treated with CHCl3 and Na OH followed by acidification salicylaldehyde is obtained. Which species are involved in the above reaction as intermediates?
Asked by Divz_slk - Sun Oct 25 01:52:24 2009 - Chemistry - 1 Answers - Comments

A. This is Reimer - Tiemann Reaction The mechanism is clearly described in the given link ( where KOH is used instead of Na OH ) H-CCl3 ( Chloroform ) + OH^- --- > CCl3^- ( Conjugate Base of Chloroform ; Trichlorocarbanion ) + H2O CCl3^- --- > :CCl2 ( Dichlorocarbene Free Radical ) + Cl^- C6H5-OH ( Phenol ) + OH^- --- > C6H5-O^- ( Phenolate anion ) + H2O C6H5-O^- + :CCl2 --- > Cl2CH-C6H4-O^- [ 2-(Dichloromethyl) phenolate ] Cl2CH-C6H4-O^- + OH^- ---> Cl(HO)CH-C6H4-O^- Cl(HO)CH-C6H4-O^- --- > O=CH-C6H4-OH ( Salicylaldehyde ) + Cl^- ( There are several intermediates , choose the ones that you want ) . .
Answered by Real Chemist - Sun Oct 25 03:10:26 2009

what is the concentration of Na OH in the diluted solution?
Q. If 65 m L of a 0.221 M Na OH is diluted to a final volume of 135 m L, what is the concentration of Na OH in the diluted solution?
Asked by Katie - Tue Dec 8 19:11:32 2009 - Chemistry - 1 Answers - Comments

A. Use the dilution equation C1V1 = C2V2 C1 = initial concentration = 0.221 M V1 = initial volume = 65 ml C2 = final concentration = ? V2 = final volume = 135 ml C2 = C1V1 / V2 = 0.221 M x 65 ml / 135 ml = 0.106 M (3 sig figs)
Answered by Lexi R - Tue Dec 8 21:06:14 2009

What would happen to the p H value of Na OH if dissolved in water?
Q. If you dissolved Na OH in water, it dissociates into Na+ and OH-. What would happen to the p H value of the solution?
Asked by JK - Sun Jan 16 22:00:01 2011 - Chemistry - 1 Answers - Comments

A. It would be a perfect 14 since Na OH is a strong base and disassociates completely
Answered by Anthony - Sun Jan 16 22:02:32 2011

How do you standardize Na OH to use it to titrate an unknown acid?
Q. I have to do an experiment where I have to figure out what is the unknown diprotic acid is (H2X) by titrating it with Na OH. But first, I have to standardize the Na OH solution. How do you do that?
Asked by M - Sat May 15 12:18:36 2010 - Chemistry - 1 Answers - Comments

A. Na OH is standardized using a primary acid such as potassium hydrogen phthalate ( KHP) dissociation of KHP : KHC8H4O4 = H+ KC8H4O4- neutralization reaction : H+ + OH- = H2O phenolphthalein in acidic solution is colorless but turns pink when all H+ ions have neutralized
Answered by Dr.A - Sat May 15 13:25:55 2010

How many moles of Na OH were added to the reaction flask?
Q. 15.50 m L of a 0.1075 M Na OH solution are required to neutralize 15.00 m L of a sulfuric acid solution. How many moles of Na OH were added to the reaction flask?
Asked by mauriciohernandez26 - Thu Mar 6 18:45:01 2008 - Chemistry - 1 Answers - Comments

A. 0.1075 moles/L times 0.1550 L is 0.0166625 moles of Na OH or 1.67 X 10 ^ (-2) moles
Answered by People A - Thu Mar 6 19:00:23 2008

What is the density of the concentrated Na OH given the following?
Q. Concentrated Na OH is 19.4 M and 50.5% Na OH by mass. The answer choices are a. 0.392 g/m L b. 1.26 g/m L c. 1.54 g/m L d. 2.55 g/m L e. 9.80 g/m L Please explain how you got the answer too, thanks!
Asked by mermaid - Thu Aug 4 01:23:20 2011 - Chemistry - 3 Answers - Comments

A. Given: 50.5 g Na OH is dissolved in 100 g solution implies 50.5 g Na OH is associated with 49.5 g water The strength of this solution is 19.4 M or 19.4 moles/liter of solution FW(Na OH) = 23 + 16 + 1 = 40 50.5 g Na OH = 50.5/40 =1.2625 moles Let the volume of the solution be V L Then 1.2625/V = 19.4 OR V = 1.2625/19.4 = 0.0651 L OR 65.1 m L The density is mass/volume = [50.5 g (Na OH) + 49.5 g (H2O)]/65.1 = 100/65.1 = 1.536 ~ 1.54 g/m L or (c)
Answered by bhola - Thu Aug 4 01:49:44 2011

Magnesium suflate is added to 456ml of 0.040 M Na OH ntil a precipitate just forms. How many grams of Mg SO4 wer?
Q. Magnesium suflate is added to 456ml of 0.040 M Na OH ntil a precipitate just forms. How many grams of Mg SO4 were added? Assume that the volume of the solution is not changed significantly by the addition of Mg So4.
Asked by Melanie - Mon Aug 2 21:26:26 2010 - Chemistry - 1 Answers - Comments

A. Magnesium suflate is added to 456ml of 0.040 M Na OH ntil a precipitate just forms. How many grams of Mg SO4 were added? Assume that the volume of the solution is not changed significantly by the addition of Mg So4. The reaction equation is shown below! Mg SO4 + Na OH Na2SO4 + Mg(OH)2 Need 2 Na and 2 OH on left side Mg SO4 + 2 Na OH Na2SO4 + Mg(OH)2 The balanced equation above states that 1 mole of Mg SO4 will react with 2 moles of Na OH to produce 1 mole of Na2SO4 and 1 mole of Mg(OH)2 456 ml of 0.040 M Na OH = 0.456 liters * 0.040 moles/ liter = 0.01824 moles of Na OH 1 mole of Mg SO4 will react with 2 moles of Na OH, so x moles of Mg SO4 will react with 0.01824 moles of Na OH. The ratio of moles of Mg SO4 to moles of Na OH = 1 : 2. So,… [cont.]
Answered by electron1 - Mon Aug 2 23:09:33 2010

How do you find the volume of Na OH dispensed?
Q. The buret reading of Na OH initial is 1.1ml the final is 10.5ml and the mol of Na OH is 0.16Na OH. Also how do you find the molar concentration.
Asked by Christina Parker - Thu Mar 24 22:55:08 2011 - Chemistry - 2 Answers - Comments

A. so, your buret is upside-down? delta V = 9.4ml 0.16moles Na OH / 0.0094L = 17M. really really strong, hardly think that this is what was actually used. this would burn a hole through your hand. check your volumes and if 0.16 = moles or molarity. molarity = molar concentration
Answered by Caroline Miller - Fri Mar 25 00:20:27 2011

What mass of hydrogen gas is produced if 17.54g Na OH is produced by the reaction?
Q. The reaction of sodium and water produces sodium hydroxide (Na OH) and hydrogen gas. How do I do this problem? What is the chemical formula of the known substance? What is the chemical formula for the unknown sunbstance? And what is the mole ratio? I really need help! Thank you!
Asked by Savannah - Sun Feb 20 20:00:56 2011 - Chemistry - 2 Answers - Comments

A. first off, u need to write out the chemical equation for the reaction. so its sodium (Na) and water (H20) forming sodium hydrozide (Na OH) and Hydrogen gas (H2 since its a diatomic) so u get this: Na + H2O ---> Na OH + H2 then u need to balance it... and you get this: 2 Na + 2H2O ---> 2Na OH + H2 to find the mass of H2 produced from 17.54 g of Na OH, use dimensional analysis take 17.54 g of Na OH and convert it to moles of Na OH by dividing by the molar mass of Na OH (40 g) then multiply this by the mole ratio (1 mol Na OH / 1 mol H2). this will get you mols of H2 produced then multiply this by the molar mass of H2 (2.02 g) ur answer will be 0.877 g H2 as far as the chem formulas for ur known and unknown substances, im not sure what… [cont.]
Answered by ~youaregurl~ - Sun Feb 20 20:15:02 2011

How do i determine a concentration of Na OH using titration with KHP if I only know volume of Na OH?
Q. I have 3 volumes, 304.1675ml , 21.2905m L, and 14.39239 m L of Na OH. I also have the mass of KHP used in each trial.
Asked by BoBo - Sun May 1 21:19:14 2011 - Chemistry - 1 Answers - Comments