Does no one know how to work a stoichiometry problem?!?
Q. I don't know where to start!! How many grams of potassium hydrogen phthalate (KHP, 204.23g/mol) can be titrated with 13.75 m L of a 0.199 M Na OH solution? KHP + Na OH ==> Na KP + H2O
Asked by :D - Sun Mar 28 18:56:03 2010 - Chemistry - 1 Answers - Comments

A. 13.75 m L * 0.199 M = 2.74 mmoles Na OH 2.74E-3 moles = x/204.23 x = 0.560 g KHP.
Answered by deadfishfactory - Sun Mar 28 19:03:04 2010

How many grams were used in this titration?
Q. Potassium hydrogen phthalate (abbreviated as KHP) has the molecular formula KHC8H4O4 and a molar mass of 204.22 g/mol. KHP has one acidic hydrogen. A sample of KHP is dissolved in 34.86 m L of water and titrated to the equivalence point with 20.88 m L of a 0.1617 M Na OH solution. How many grams of KHP were used in the titration? KHC8H4O4(aq) + Na OH(aq) ==> KNa C8H4O4(aq) + H2O(l)
Asked by - Tue Feb 9 02:01:47 2010 - Chemistry - 3 Answers - Comments

A. From the balanced equation: 1mol KHP reacts with 1mol Na OH Mol Na OH in 20.88ml 0.1617M Na OH: 20.88/1000*0.1617 = 0.003376mol Na OH Therefore you used 0.003376mol KHP Molar mass KHP = 204.22g/mol 0.003375mol = 0.003376*204.22 = 0.6895g KHP was used.
Answered by Trevor H - Tue Feb 9 09:56:54 2010

help with chemistry questions?
Q. you need to determine the concentration of a solution of HBr(aq). at your disposal you have a bottle of an unknown concentration of aqueous KOH, a bottle of crystalline KHP, and lots of distilled water a)KHP (potassium hydrogen phlalate, 204.23 g/mol) is an acid that is used to determine the concentration of bases. how would you prepare 500 ml of a standard solution 0.250 M KHP? b) you react 33.47 ml of the 0.250 M solution of KHP with 25.06 ml of the unknown KOH(aq) solution: KHP(aq) + KOH(aq) -> K2P(aq) + H2O what is the molar concentration of the "unknown" KOH solution c) you now react the KOH solution of known concentration with the unknown solution of HBr. write out the neutralization reaction that takes place d) if you… [cont.]
Asked by qwert.tyui - Wed Jun 2 21:48:15 2010 - Chemistry - 1 Answers - Comments

A. moles KHP = 0.500 L x 0.250 M= 0.125 mass KHP required = 0.125 mol x 204.23 g/mol=25.5 g moles KHP = 0.03347 L x 0.250 M= 0.00837 = moles KOH molarity KOH = 0.00837/ 0.02506 L=0.334 M KOH + HBr = KBr + H2O moles KOH = 0.334 M x 0.0427 L=0.0143 Molar concentration HBr = 0.0143 / 0.02508 L=0.570 M p H = - log 0.570=0.244
Answered by Dr.A - Thu Jun 3 09:01:10 2010

Potassium hydrogen phthalate, KHP Acids & Bases Q
Q. Potassium hydrogen phthalate, KHP, (molar mass = 204.2 g/mol)is one of the most commonly used acids for standardizing solutions containing bases. KHP is a monoprotic weak acid with Calculate the p H of the solution that results when 0.80 g of KHP is dissolved in enough water to produce 25.0 m L of solution. a. 3.11 b. 4.60 c. 5.41 d. 1.95 Ka = 3.91 X10 ^-6 I dont know how Ka is relevant... i thought to solve this, i needed to know the conc. of acid and use -log(acid) but that didnt get me any of the answers... Your answer of 5.41 is wrong...
Asked by monica - Mon Aug 11 07:54:32 2008 - Chemistry - 2 Answers - Comments

A. first determine the molarity of the KHPH = 0.8 grams/0.2042/25 m L = 0.1567 M (M = mmoles/m L) this amount of KHPH will ionize a small amount ( x ) to yield X amounts of H+ and X amount of the base so based on Ka [ x ][x]/ 0.1567-x = 3.91 X 10^-6 X^2 = 1.567X 10^-1 x 3.91 X 10^-6 x is small compared to 0.1567 M and is neglected solving for x x= 7.84 X 10^-4 = [H+] so p H = -og 7.84X 10^-4 = 3.114 = a
Answered by Merlin's Feline - Mon Aug 11 09:15:16 2008

need more chemistry help!?
Q. 1) A 5.00g mixture of sodium chloride and sodium carbonate Na2CO3(s) is heated to decompose the sodium carbonate into sodium oxide and carbon dioxide by the following reaction: Na2CO3(s) ==> Na2O(s) + CO2(g). If the original mixture is 25.1% sodium carbonate, what is the mass of the final mixture after heating? 2) How many grams of potassium hydrogen phthalate (KHP, 204.23g/mol) can be titrated with 13.14 m L of a 0.196 M Na OH solution? KHP + Na OH ==> Na KP + H2O. 3) In the lab you used 0.729 g Cu Cl2*2H2O in the process of determing the chemical formula of this copper chloride salt. How much aluminum (Al) was lost from the coil when isolating the solid copper metal? The redox reaction between copper II ions and aluminum is: 3 Cu2+… [cont.]
Asked by :D - Fri Mar 26 14:48:19 2010 - Chemistry - 1 Answers - Comments

A. What you don't understand is how chemists express weight and concentration. Thery use a number called a mole. This many atoms of any element equals its atomic weight in the periodic table. Molecular weight is the sum of atomic weight. Concentration is expressed in terms of moilarity. The units are moles/liter. To solve most chemistry problems: 1.) express all the weights given in terms of moles. 2.) write a balanced chemical equation describing the reaction. A balanced equation has equal numbers of atoms on both sides. The number of molecules don't have to be equal, though. The numbers of molecules is the same as the number of moles, or fractions of moles. This applies to the ratios of molecules as well. Since we are also dealing with… [cont.]
Answered by Roger S - Fri Mar 26 14:55:30 2010

Lab excercise calculating molarity of Na OH solution?
Q. The following data were collected during the titration of a solid sample of potassium acid phthalate ("KHP") with a solution of sodium hydroxide. The Na OH solution was added to the KHP from a buret. From the experimental data below, calculate the molarity of the Na OH solution. Data: Molar Mass of potassium acid phthalate= 204.2 g/mol Mass of weighing bottle plus KHP = 23.5844 g Mass of weighing bottle= 22.1831 g Initial buret reading = .39 m L Final Buret reading = 37.12 m L
Asked by Olivia:) - Wed Oct 22 09:29:28 2008 - Chemistry - 1 Answers - Comments

A. mass KHP = 23.5844 - 22.1831 = 1.401 g moles KHP = 1.401 g / 204.2 = 0.006862 = moles Na OH volume Na OH = 37.12 - 0.39 = 36.73 m L M = 0.006862 / 0.03673 L = 0.1868
Answered by Dr.A - Wed Oct 22 12:07:28 2008

What is the molarity of Na OH? (titration question)?
Q. You titrate .4278g of potassium acid phthalate (KHP) with 22.86 m L of Na OH to reach the phenolphthalein pink endpoint. What is the molarity of the Na OH? Molar mass of potassium acid phthalate (KHP) is 204.33 g/mol. Any help on this would be really great! thanks.
Asked by thesekeys - Wed Feb 14 12:03:33 2007 - Chemistry - 2 Answers - Comments

A. 1) calculate how many moles of KHP there is in the sample. 2) at the end point, moles of Na OH added = moles of KHP you started with. 3) so, now you know moles of Na OH added, and your volume is 22.86 m L so convert into molarity (moles/L) N.B. Don't forget to change m L to L for the last step!
Answered by raerae_2001 - Wed Feb 14 12:38:48 2007

I need help in finding the initial concentration of khp?
Q. 0.4209 g of potassium hydrogen phthalate, KHP (204.23 g/ mol), in 100 ml oof water was titrated with Na OH. What was the intial concentration of the KHP? I'm not sure where to start
Asked by - Sun Apr 15 17:49:24 2007 - Chemistry - 2 Answers - Comments

A. 0.4209/(204.23 x 0.1) The 0.1 is for 0.1 litres.
Answered by Gervald F - Sun Apr 15 17:57:10 2007

Basic Chemistry Homework Help?
Q. To standardize a solution of approximately 0.1 M Na OH, a student dissolved a known mass of KHP (potassium hydrogen phthalate, KHC8H4O4, MW 204.2 g/mol) in water, and titrated it with Na OH whose molarity was only approximately known. The KHP solution was made by dissolving 0.4023g in water. It took 19.57ml of Na OH solution to titrate the KHP solution to a phenolphthalein endpoint. Molarity of the Na OH solution = ___ A total of 16.97m L of 0.1012 M Na OH was required to reach a phenolphthalein end point when titrating a 2.12m L sample of vinegar. Calculate the molarity of acetic acid present in the sample. Molarity of acetic acid = ___ A total of 16.08m L of 0.01006 M Na OH was required to reach phenolphthalein end point when titrating a… [cont.]
Asked by CMoney - Thu Oct 22 02:16:07 2009 - Chemistry - 1 Answers - Comments

A. moles KHP = 0.4023 / 204.2 =0.001970 M ( Na OH) = 0.001970/ 0.01957 L=0.1007 moles Na OH = 0.01697 L x 0.1012 M=0.001717 M ( acetic acid )= 0.001717 / 2.12 x 10^-3 L=0.8101 moles Na OH = 0.01608 L x 0.01006 M=0.0001618 moles citric acid = 0.0001618/3 =0.0005392 M = 0.0005392 / 5.02 x 10^-3 L=0.01074
Answered by Dr.A - Thu Oct 22 05:44:57 2009

Chemistry 1 Help Please?
Q. I have a big test tomorow and i don't know how to solve the following questions.can u guys like tell me how to do it and then why. thanks alot (1) Na OH+KHP->Na KP+H20 (MM of KHP=204.2g/mol),KHP isnt actually potassium+hydrogen+potassium a.how many grams of KHP are needed to react completely with 35.2m L of 0.1208M Na OH? b. Na OH is standarized by titrating KHP to the phenolphthalein end point. If 0.3265g of KHP required 14.80m L of Na OH to reach the end point, find the molarity of the Na OH solution c.If 2.156g of vinegar needed 16.54m L of 0.1120M Na OH to neutralize it, find the % acetic acid in the vinegar solution. (2)Sodium carbonate can be used as a primary standard for finding the molarity of acid solutions. If 0.6846g of… [cont.]
Asked by hyperstrike11 - Tue Apr 24 23:52:43 2007 - Chemistry - 1 Answers - Comments

A. hmm... these are pretty tough questions to explain like this but... lets do at least #1. i try not to give out answers but i'll try to explain how to do it. a. so from the given equation, you can see that for every 1 mole of Na OH, 1 mole of KHP is required. (1:1 ratio). Hope that is clear. step 1: they give you 35.2m L and 0.1208M so from those two numbers, you can figure out the MOLES of Na OH you have because we know that Molarity = Moles/Liter Remember, convert m L --> Liters. step 2: once you find the number of moles of Na OH you have, you know you are going to use up the same number of moles of KHP (because of the 1:1 ratio mentioned above). So just convert moles --> grams by using the molar mass of KHP. Moles x 204.2g/mol b.… [cont.]
Answered by Xenon - Wed Apr 25 01:32:03 2007

How do I do this without another molarity or something? What is MW?
Q. How many grams of KHP (potassium hydrogen phthalate MW = 204.2 g/mol) are needed to neutralize 17.64 m L of a 0.1001 M Na OH solution?
Asked by diamond - Sun Aug 8 23:02:02 2010 - Chemistry - 1 Answers - Comments

A. MW stands for molecular weight, which basically means that that many grams are in one mole of KHP. First, you must calculate how many equivalents of base you have. 0.01764 L*0.1001= 0.001766 mol of base, so you need 0.001766 mol of acid. Since there is 204.2 g in one mole of KHP, you would need 0.3606 g of KHP to neutralize the base. Hope this helps!
Answered by - Mon Aug 9 00:02:47 2010

Need a quick answer for an easy chemistry question?
Q. A basic solution of Ca(OH)2 was standardized versus 0.2351 g of KHP (potassium hydrogen phtalate) acid - the volume of Ca(OH)2 needed to fully titrate with this mass of KHP was 38.12 m L. What is the standardized concentration of this Ca(OH)2 solution? (Molar masses - K+: 39.10 g/mol, Hydrogen Phthalate( ): 165.12 g/mol, Ca2+: 40.08 g/mol, OH : 17.01 g/mol.) I thought that moles KHP = moles Ca(OH)2; therefore, (grams KHP/molar mass KHP) = (concentration Ca(OH)2)(volume Ca(OH)2) and solve for concentration. This gives me 0.0303M, but the answer is 0.0151M. Why? Is it because of the charge on calcium or something? Please help quickly my midterm is in 50 minutes haha.
Asked by tzier187 - Mon May 7 11:11:12 2007 - Chemistry - 3 Answers - Comments

A. Yes, it is because of the charge on calcium, in a way. Because calcium forms a Ca2+ cation, it supports 2 OH groups instead of 1. Therefore, each mole of Ca(OH)2 can neutralize two moles of KHP, and when you didn't take this into account, you calculated a molarity that was exactly twice what you should have gotten.
Answered by Amy F - Mon May 7 11:55:33 2007

AP Chemistry Question help?
Q. The molarity of an Na OH solution was de- termined by using the primary standard potassium hydrogen phthalate (KHP, formula weight = 204.2 g/mol). Na OH + KHP ! Na KP + H2O It took 20.0 m L of the Na OH to completely react with 0.450 g of KHP. What was the molarity of the Na OH? 1. 0.044 M 2. 0.088 M 3. 4.59 M 4. 0.110 M
Asked by - Mon Oct 17 21:19:33 2011 - Chemistry - 1 Answers - Comments

Chem Help?? Acid/Base Titrations?
Q. A sample of 1.005 g of KHP (Potassium Hydrogen Phthalate, molar mass = 204.22g/mol) was dissolved in ~ 25 m L distilled water and titrated with a Na OH solution of unknown concentration. If 28.68 m L of base was used to reach the endpoint, what was the concentration of the Na OH (in moles per liter)? Use correct number of significant figures when reporting answer.
Asked by Curious - Mon Oct 26 21:02:50 2009 - Chemistry - 1 Answers - Comments

Potassium hydrogen phthalate (abbreviated as KHP) has the molecular formula KHC8H4O4 and a molar mass of?
Q. Potassium hydrogen phthalate (abbreviated as KHP) has the molecular formula KHC8H4O4 and a molar mass of 204.22 g/mol. KHP has one acidic hydrogen. A sample of KHP is dissolved in 50.63m L of water and titrated to the equivalence point with 13.00 m L of a 0.3315 M Na OH solution. How many grams of KHP were used in the titration? KHC8H4O4(aq) + Na OH(aq) ===> KNa C8H4O4(aq) + H2O(l)
Asked by Lisa M - Wed Jul 2 02:20:13 2008 - Chemistry - 2 Answers - Comments

A. HP- + OH- ---> P2- + H2O moles of OH- = M*V = 0.3315 * 0.013 = 4.310*10^-3 mol = moles of KHP grams of KHP = moles * MM = 4.310*10^-3 * 204.22 = 0.880 g
Answered by - Wed Jul 2 05:07:48 2008

Chemistry acid base titration question help?
Q. A sample of 1.018 g of KHP (Potassium Hydrogen Phthalate, molar mass = 204.22g/mol) was dissolved in ~ 25 m L distilled water and titrated with a Na OH solution of unknown concentration. If 29.61 m L of base was used to reach the endpoint, what was the concentration of the Na OH (in moles per liter)?
Asked by Dhananjay K - Thu Oct 21 05:47:31 2010 - Chemistry - 2 Answers - Comments

A. C1V1 =C2V2 no of moles of KHP =1.018/204.22 = 0.005 mol molarity =mol of solut /vol(L) = 0.005/(25/1000) =0.2 mol/L 0.2 * 25 = C2 * 29.61 C2 =0.169 mol/L Na OH Conc.
Answered by - Thu Oct 21 06:04:41 2010

Calculate p H of the solution that results when 0.40 g of KHP is dissolved in water to produce 25.0ml solution.?
Q. potassium hydrogen phthalate (KHP): molar mass = 204.2 g/mol KHP: monoprotic weak acid with a Ka=3.91x10^-6
Asked by Muppet D - Wed Mar 11 12:53:14 2009 - Chemistry - 1 Answers - Comments

A. moles KHP = 0.40 g/ 204.2 g/mol=0.00196 initial concentration = 0.00196/ 0.0250 L=0.0784 M 3.91 x 10^-6 = x^2 / 0.0784-x x = [H+]= 0.000554 M p H =3.26
Answered by Dr.A - Wed Mar 11 13:37:05 2009

Calculate the Molarity of Sodium Hydroxide?
Q. If 16.95m L of a solution of Na OH is required to titrate a 0.3556g sample of KHP (potassium hydrogen phthalate) MW - 204.2g/mol. What is the molarity of the sodium hyroxide solution?
Asked by - Wed Feb 24 22:39:28 2010 - Chemistry - 1 Answers - Comments

A. ANSWER: 0.1027M Na Oh solution: Find mole of KHP: mole KHP = 0.3556 g * (1 mole KHP/204.2 g) = 0.001741 KHP is monoprotic, so 1 mole of KHP would be neutralized by 1 mole Na OH. so, mole Na OH = 0.001741 === Volume of Na OH = 16.95 m L (1 L/1000 m L) = 0.01695 L Molarity of Na OH = mole Na OH/volume of Na OH in liter = 0.001741 / 0.01695 L = 0.1027M Na OH
Answered by - Thu Feb 25 00:39:05 2010

How to find p H of KHP titration at equivalence point?
Q. " Potassium hydrogen phthalate, known as KHP (molar mass = 204.22 g/mol), can be obtained in high purity and is used to determine the concentration of solutions of strong bases by the reaction: HP- + OH- -> H2O + P2- If a typical titration experiment begins with about 0.5 g KHP and has a final volume of about 100 m L, what would be an appropriate indicator to use? The p Ka for HP- is 5.51. " I believe I have to find the p H at equivalence point but I am unsure as to how to do that. Then I can just use a chart and determine which indicator to use.
Asked by - Fri Oct 22 16:46:27 2010 - Chemistry - 1 Answers - Comments

A. find the Ka for KHP if the p Ka = 5.51 then, Ka = 10^-5.51 Ka = 3.09 e-6 (I will pooint out that alot of other Ka's for KHP are out there, like 3.9 e-6) === find moles, using molar mass: 0.5 g KHP @ 204.22 g/mol), = 0.00245 moles KHP by the equation: HP- + OH- -> H2O + P2- 0.00245 moles KHP produces 0.00245 moles P-2 find molarity of that 100 ml solution: 0.00245 moles P-2 / 0.100 litre = 0.0245 Molar P-2 === P2- is a strong conjugate base of the weak acid KHP it will do a hydrolysis at the equivalence point making the solution basic: P2- & H2O --> HP- & OH- Khydrolysis = Kwater / K acid = (1e-14) / (3.09 e-6) = 3.236 e-9 Kh = [HP-] [OH-] / [P2-] 3.2 e-9 = [X] [X] / [0.0245] X2 = 7.93 e-11 X = {OH-] = 8.9 e-6 p OH = 5.05… [cont.]
Answered by Steve O - Fri Oct 22 20:05:38 2010

It takes 45.78 m L of the Na OH solution to neutralize 1.256 grams of KHP. what is the molarity of ...?
Q. a sodium hydroxide solution of unknown concentration is standardized against potassium biphthlate. it takes 45.78 m L of the Na OH solution to neutralize 1.256 grams of KHP. what is the molarity of the sodium hydroxide? molar mass of KHP = 204 g/mol can someone break down the steps for me? I have a few problems like this and if i can see how its broken down it will be easier for me to solve the next few. thanks for your time.
Asked by OhSnap05 - Wed Feb 11 12:05:59 2009 - Chemistry - 1 Answers - Comments

A. moarity is 1.193.
Answered by avinash - Wed Feb 11 12:20:03 2009