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 - 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

If 25 m L of 2.5 M HCl are titrated with 2.2 M Na OH, what is the p H after adding 15 m L Na OH?
Q.
Asked by ILoveMaths07 - Sat May 28 15:11:10 2011 - Chemistry - 3 Answers - Comments

A. Equation: Na OH+ HCl Na Cl + H2O 1mol Na OH reacts with 1 mol HCl Mol HCl in 25m L of 2.5M HCl solution = 25/1000*2.5 = 0.0625 mol HCl Mol Na OH in 15m L of 2.2M Na OH = 15/1000*2.2 = 0.033 mol Na OH When reacted there is 0.0625-0.033 = 0.0295 mol HCl remaining unreacted, dissolved in 40m L solution. Molarity of HCl solution = 0.0295/0.040 = 0.7375M HCl because HCl is a strong acid: [H+] = [HCl] = 0.7375M p H = - log [H+] p H = - log 0.7375 p H = 0.13 p H of the final solution = 0.13
Answered by Trevor H - Sat May 28 15:32:33 2011

You are given solutions of HCl and Na OH and must determine their concentrations.?
Q. You use 30.4 m L of Na OH to titrate 100. m L of HCl and 17.8 m L of Na OH to titrate 50.0 m L of 0.0750 M H2SO4. Find the molarity of HCl. I found the molarity of Na OH, but I'm having problems with HCl. Work shown would be nice.
Asked by - Mon Nov 1 21:52:06 2010 - Chemistry - 1 Answers - Comments

A. moles H2SO4 = 0.0500 L x 0.0750 M=0.00375 moles Na OH = 0.00375 x 2 =0.00750 M of Na OH = 0.00750 / 0.0178 L=0.421 moles Na OH = 0.421 M x 0.0304 L=0.0128 moles HCl = 0.0128 molarity HCl = 0.0128/ 0.100 L = 0.128 M
Answered by Dr.A - Tue Nov 2 09:39:28 2010

A 50.00-m L sample of 0.500 M HCl is titrated with 0.500 M Na OH. What is the p H of the solution after 43.82 m L?
Q. A 50.00-m L sample of 0.500 M HCl is titrated with 0.500 M Na OH. What is the p H of the solution after 43.82 m L of Na OH have been added to the acid?
Asked by - Fri Jun 25 13:34:37 2010 - Chemistry - 1 Answers - Comments

A. moles HCl = 0.05000 L x 0.500 M=0.0250 moles Na OH = 0.04382 L x 0.500 M= 0.0219 moles HCl in excess = 0.0250 - 0.0219 =0.00310 total volume = 50.00 + 43.82 = 93.82 m L = 0.09382 L [H+] = 0.00310 / 0.09382 L =0.0330 M p H = - log 0.0330 =1.48
Answered by Dr.A - Fri Jun 25 14:52:22 2010

When 30 ml of 1 M HCl are titrated with 0.5M Na OH what is its p H after adding 0, 1, 30, 60, and 75 ml base?
Q. Any help is appreciated. Thank You.
Asked by SpeechD - Wed Apr 9 17:47:26 2008 - Chemistry - 1 Answers - Comments

A. Initial concentration HCl = 1 M p H = 0 Moles HCl = 0.030 L x 1 M = 0.030 Moles OH- = 0.001 L x 0.5 M = 0.0005 Moles H+ in excess = 0.030 - 0.0005 = 0.0295 total volume = 31 m L = 0.031 L [H+] = 0.0295 / 0.031 = 0.952 M p H =0.0215 moles OH- = 0.030 L x 0.5 M = 0.015 Moles H+ in excess = 0.030 - 0.015 = 0.015 total volume = 30 + 60 = 60 m L = 0.060 L [H+] = 0.015 / 0.060 =0.25 M p H = 0.602
Answered by Dr.A - Thu Apr 10 04:41:37 2008

What is the molarity of an HCl solution if 12.0 m L HCl solution is titrated with 30.6 m L of 0.165M Na OH?
Q. HCl(aq) + Na OH(aq) to Na Cl(aq) + H20(l) ???molarity=???
Asked by - Wed Nov 9 17:01:56 2011 - Chemistry - 1 Answers - Comments

100m L of 0.200M HCl is titrated with 0.250M Na OH. What is the p H of the solution at equivalence point?
Q.
Asked by Ari - Wed Mar 4 10:28:02 2009 - Chemistry - 1 Answers - Comments

A. The equivalence point will be when the amount of Na OH is equal to the amount of HCl, and the p H should be neutral, or 7 because the reaction is HCl + Na OH --> Na Cl + H2O
Answered by Simonizer1218 - Wed Mar 4 11:01:13 2009

4. A 50.0 m L sample of 0.1084 M HCl was titrated with a solution of Na OH and 28.4 m L of the titrant were used.?
Q. 4.A 50.0 m L sample of 0.1084 M HCl was titrated with a solution of Na OH and 28.4 m L of the titrant were used. What is the molar concentration of the Na OH solution? Cause I need to learn how ... thank you very much,
Asked by - Wed Oct 19 23:08:22 2011 - Chemistry - 1 Answers - Comments

A 25.00 m L sample of a 0.2158 M solution of Na OH is titrated with 16.88 m L of HCl. What is the concentration o?
Q. What is the concentration of HCl solution?
Asked by Neisha - Thu Dec 3 18:56:38 2009 - Chemistry - 1 Answers - Comments

A. m1v1=m2v2 Na OH 25m L=v1 Na OH 0.2158M=m1 HCl 16.88m L=v2 hope this helps :)
Answered by - Thu Dec 3 19:21:46 2009

.5 g of Na HCO3 treated with 40 ml HCL. Excess HCL back titrated with 9.5 ml Of .31 M Na Oh. Wht is M of HCL?
Q. given mmol HCL = mmol NAHCO3 + mmol Naoh
Asked by om - Tue May 5 12:19:52 2009 - Chemistry - 2 Answers - Comments

A. for total reaction mmoles Na HCO3 = mmoles of HCl mmoles of Na HCO3 = 500mg/84mg/mmol = 5.952 mmoles soo 5.952 mmoles = total mmoles HCl - ( 9.5 x0.31 M ) total mmoles HCL = 5.952+ 2.945 total mmoles HCl = 8.897::: molarity = mmoles/m L = 8.897 mmol/40m L = 0.2224M
Answered by Merlin's Feline - Tue May 5 14:23:49 2009

100 m L of 0.200 M HCl is titrated with 0.250 M Na OH.?
Q. INFO: 100 m L of 0.200 M HCl is titrated with 0.250 M Na OH. PART A: What is the p H of the solution after 50.0 m L of base has been added? PART B: What is the p H of the solution at the equivalence point?
Asked by - Sat Nov 7 00:49:54 2009 - Chemistry - 1 Answers - Comments

A. moles HCl = 0.100 L x 0.200 M = 0.0200 moles Na OH = 0.0500 L x 0.250 M=0.0125 moles HCl in excess = 0.0200 - 0.0125 =0.00750 total volume = 0.150 L [H+]= 0.00750/ 0.150L=0.0500 M p H = - log 0.0500=1.30 at the equivalence point moles H+ = moles OH- so p H = 7.00
Answered by Dr.A - Sat Nov 7 05:02:31 2009

Titration with HCl and Na OH?!?!?!?
Q. You are given solutions of HCl and Na OH and must determine their concentrations. You use 27.8 m L of Na OH to titrate 100. m L of HCl and 22.2 m L of Na OH to titrate 50.0 m L of 0.0812 M H2SO4. Find the unknown concentrations.
Asked by LovesDierks. - Tue Nov 4 23:21:35 2008 - Chemistry - 2 Answers - Comments

A. 2Na OH + H2SO4 --> Na2SO4 + 2H2O 50.0 m L x 0.0812 M H2SO4 = 4.06 mmol H2SO4 and 8.12 mmol H^+1 22.0 m L Na OH x ? M = 8.12 mmol Na OH Na OH molarity = 0.369 Na OH + HCl --> Na Cl + H2O 27.8 m L x 0.369 M Na OH = 10.26 mmol Na OH 100. m L HCl x ? M = 10.26 mmol HCl HCl molarity = 0.1026 M (0.103 M)
Answered by skipper - Tue Nov 4 23:33:24 2008

A 100 ml solution of 0.1 N HCL was titrated with 0.2 N Na OH solution.The titration was discontinued?
Q. after adding 30 ml of Na OH solution.The remaining titration was completed by adding 0.25 N KOH solution.The volume of KOH required for completing the titration is: A)70 ml B)32 ml C)35 ml D)16 ml please give me the full solution.thnx in advance
Asked by anurag D - Fri Jun 26 09:49:13 2009 - Chemistry - 6 Answers - Comments

A. For HCl, KOH and Na OH normality = molarity moles HCl = 0.100 L x 0.1 N = 0.01 Moles OH neeed = 0.01 moles Na OH added = 0.030 x 0.2 = 0.006 moles KOH needed = 0.01 - 0.006=0.004 V = 0.004/0.25=0.016 L => 16 m L
Answered by Dr.A - Sat Jun 27 12:51:47 2009

a 25.0m L sample of HCL was titrated to the endpoint with 15.0m L of 2.0 N Na OH what is the normality of HCL?
Q.
Asked by juliarkempke - Thu May 8 11:20:50 2008 - Chemistry - 1 Answers - Comments

A. at the equivalence point N*V HCl = N*V Na OH N HCl = N*V Na OH / V HCl = 2.0*15.0 / 25.0 = 1.20 N
Answered by - Thu May 8 11:59:58 2008

The titration of HCl with Na OH is represented by the equation HCl+Na OH--->Na Cl+H20.?
Q. What volume of 0.100M HCl is required to titrate 50.0m L of 0.500M Na OH?
Asked by brissy - Sun Jun 27 00:22:50 2010 - Chemistry - 1 Answers - Comments

A. C1V1 = C2V2 0.100 M * V1 = 50.0 m L * 0.500 M V1 = 250 m L
Answered by deadfishfactory - Sun Jun 27 00:25:06 2010

Na OH, HCl Titration question?
Q. I was doing a titration lab, and after 6-8 attempts I could not get the desire results so this is why I am asking for help So this is what I know: The HCl used has a Molality of: 0.010M HCl I use Na OH as the base, and have to find the approximate ml used to Titrate, With that I could find the rest. So what I need to exactly know: How many ml of Na OH is needed to titrate 25.0ml HCl. The Process used in the lab (if needed): Basically I would take 25.0ml HCl put a drop of Phenolphthalein Indicator into it. Then separately take some, for example 25.0ml Na OH base in a grad cyl. Then the HCl and indicator would go below the buret while the Na OH is poured into it. Then I would let the Na Oh go into the HCl indicator mix in drops while… [cont.]
Asked by - Mon Jun 6 18:05:11 2011 - Chemistry - 1 Answers - Comments

When a 1M solution of HCl is titrated with 1M solution of Na OH what happens to the solution's conductivity?
Q. A. The conductivity will decrease B. The conductivity will increase C. The conductivity will remain the same D. There will be no conductivity Please Help!
Asked by Brett P - Mon Jun 9 10:24:40 2008 - Chemistry - 4 Answers - Comments

A. Thing to remember here: "conductivity of a solution is proportional to its ion concentration" A. You started with 1M HCl and it dissociates completely (high conductivity) . After titration, the acid and base react completely (no Na OH or HCl left), producing water, and the same number of moles of Na Cl as the original HCl. (write down balanced equation to see this.) Na Cl also dissociates completely. However, because the reaction also produced water, this means the molarity increased for Na Cl also (same number of moles of Na Cl in a greater amount of water). This means a lower concentration of Na Cl than the original HCl, leading to a decrease in conductivity. :)
Answered by Cesium - Mon Jun 9 10:35:46 2008

How to find molarity of HCl and Na OH after titration?
Q. Okay so in a lab I took 25 m L of HCl and titrated it with Na OH, it took 36.29 m L of Na OH to neutralize the solution. (When it changed color due to the indicator) Now I need to find the molarity of both the HCl and Na OH, I know the formula for molarity but I don't know how to do it if the molarities of both the acid and base are unknown. Any help? Yeah but my lab sheet doesn't give the molarity of either one, I just have the m L used in the titration.
Asked by - Mon Oct 3 21:55:57 2011 - Chemistry - 1 Answers - Comments

45 m L of Na OH is titrated with 0.14 M HCl. It takes 33 m L of HCl to reach the equivalence point. What is the c?
Q. When you add an acid to a buffer, how does the buffer maintain the p H of the solution? by reacting with the hydroxide ions by adding more hydrogen ions to the solution by reacting with the hydrogen ions by reacting with the conjugate base 2. What is needed to make a buffer solution? an acid and its conjugate base an acid and a base two bases two salts 3. Which of the following reactions is titration an example of? endothermic exothermic neutralization buffering 4. 45 m L of Na OH is titrated with 0.14 M HCl. It takes 33 m L of HCl to reach the equivalence point. What is the concentration of the Na OH solution? 0.19 M 0.14 M 0.10 M 0.08 M 5. 105 m L of HCl is titrated with 0.10 M Na OH. If the equivalence point is reached when 75 m L of Na… [cont.]
Asked by I love my tree - Fri Mar 5 01:22:53 2010 - Chemistry - 1 Answers - Comments

A. 45 m L of Na OH is titrated with 0.14 M HCl. It takes 33 m L of HCl to reach the equivalence point. What is the c? When you add an acid to a buffer, how does the buffer maintain the p H of the solution? Acid + water Hydronium ion + negative ion of acid HCl + H2O H3O^+1 + Cl^-1 The buffer attracts an H+1 off the H3O+1 ion by reacting with the hydroxide ions by adding more hydrogen ions to the solution by reacting with the hydrogen ions yes The buffer attracts a proton (H^+1 ion) off the Hydronium ion. Since the Hydrogen atom has 1 proton and 1 electron, a H^+1 ion would be a proton. The proton can not exist in alone. It needs an electron to shield it. I used to tell my students that Hydrogen ions are unshielded protons. All atoms and… [cont.]
Answered by electron1 - Fri Mar 5 02:07:34 2010

Which of the following titrations result in an acidic solution at the equivalence point?
Q. Here ar ethe choices: CH3COOH titrated with Na OH KF titrated with KOH HCl titrated with Na OH C5H5N titrated with HCl I think C5H5N when titrated with HCI will give the acidic solution equivalence point. Does anyone know if this is right??? Thank you:)
Asked by Rose - Wed Nov 12 19:25:27 2008 - Chemistry - 1 Answers - Comments

A. Yupp, its correct. This is because HCl is an acid so as you add acid your solution becomes more acidic towards the equivalence point.
Answered by makemesmillle:] - Wed Nov 12 19:29:00 2008