Back titration? Find the neutralizing power of the antacid in terms of mmol H+/g of antacid?
Q. 1.3188g of antacid is weighed and mixed with 75.00 mL of excess 0.1746 M HCl. The excess acid required 27.20 mL of 0.09767 M NaOH for back titration. Calculate the neutralizing power of the antacid in terms of mmol H+ per gram of antacid.
Asked by Agatha S - Tue Sep 30 22:14:00 2008 - - 1 Answers - 0 Comments

A. Moles NaOH = 0.02720 L x 0.09767 M = 0.002657 Moles HCl = 0.07500 L x 0.1746 = 0.01310 Moles HCl used to titrate the antiacid = 0.01310 - 0.002657 =0.009443 = 9.443 mmol 1.3188 : 9.443 = 1 : x x = 7.160 mmol per gram
Answered by Dr.A - Wed Oct 1 15:10:29 2008

Chemistry qn - how to get mmol from moles -10 points!?
Q. Ok, I did this experiment where we had varoius volumes of h20, Ca(oh)2 and HCL to add. It was a ph experiment. So e.g container 4 had 40ml H20, 10ml of Ca(OH)2, 0ml Hcl, and the blank pH for this was 11.88 (I don't know wether that is relevant bt I added it just in case). Now I need to work out how much mmol OH- was added and how much mmol H+ was added. I'nve got a hole column of these to do but I can't for the life of me remmber how to do it! (I did this chemistry two years ago!) Anyway, any good help gets you 10 points straight up. Thanks!
Asked by Beep? Beep Shmeep - Sun Mar 22 21:11:49 2009 - - 2 Answers - 0 Comments

A. Molarity is moles/liter AND is also mmoles/mL So # mmoles = (molarity)(volume in mL) neat, aye?
Answered by Doc - Sun Mar 22 23:52:52 2009

are the final concentrations of NH3(aq) and NH4Cl(aq), respectively?
Q. If 30.0 mmol HCl(g) is added to 1.00 L of a buffer that is 0.340 M NH3(aq) and 0.290 M NH4Cl(aq), what are the final concentrations of NH3(aq) and NH4Cl(aq), respectively? Assume no volume change. a. 0.310 M and 0.320 M b. 0.310M and 0.290 M c. 0.370 M and 0.290 M d. 0.370 M and 0.320 M
Asked by Tina L - Thu Mar 6 22:48:30 2008 - - 1 Answers - 0 Comments

A. moles NH3 = 0.340 moles NH4+ = 0.290 moles H+ = 0.0300 NH3 + H+ >> NH4+ moles NH3 = 0.340 - 0.0300 = 0.310 moles NH4+ = 0.290 + 0.030 = 0.320 Since the volume is 1 L M = 0.310 and M = 0.320 a is the answer
Answered by Dr.A - Fri Mar 7 09:28:41 2008

Strong acid-Strong Base Titration?
Q. 1.0560 g of antiacid is weighed and mixed with 75.00 mL of excess 0.1126 M HCl. The excess acid required 4.68mL of 0.1008 M NaOH for back titration. Calculate the nuetralizing power of the antacid in terms of mmol H+ per gram of antacid. For the calculations the # of milimoles of antacid in the sample is deteremined by subtraction the # of mmol of the excess of HCl, nuetralized by the standardized NaOH, from the total # of mmol of HCl added to the antacid at the start of the experiment. if you could please answer this question it would be greatly appreciated. Thanks
Asked by Aphrodisiac - Wed Oct 3 00:07:19 2007 - - 1 Answers - 0 Comments

A. moles NaOH = 4.68 x 0.1008 /1000 = 0.000472 Moles HCl = 75 x 0.1126 /1000 = 0.00845 Moles HCl needed to neutralize the antiacid = 0.00798 1 : x = 1.0560 : 0.00798 x = 0.007555 mol H+ per gram = 7.555 mmol
Answered by Dr.A - Sat Oct 6 09:21:16 2007

Reactions in aqeous Solns.?
Q. 1. One antacid contains 334 mg of active ingredient, NaAl(OH)2CO3. Calculate number of mmol of HCL that the table can neutralize. The Neutralization produces NaCl, AlCl3, and H2O. 2. sample of 70.5 of potassium is added to 15ml of .050M silver nitrate. Calculate the theoretical yield of the precipitate that formed. 3. What volume of .500M BaCl2 is required to completely react with 4.32g of Na2SO4? For each of these i now that i will need to find the limiting reactant, but whats next? 2. 70.5 mg of potassium phosphate
Asked by wqel12 - Thu Apr 26 01:52:05 2007 - - 1 Answers - 0 Comments

A. NaAl(OH)2CO3 + 4HCl -> NaCl + AlCl3 + 3H2O + CO2 Molar mass of the antacid = 144 g millimol = 334/144 = 2.319 millimol of HCl = 4x2.319 = 9.276 2.) 70.5 ??? 3) BaCl2 + Na2SO4 -> BaSO4 + 2NaCl Molar mass of Na2SO4 = 142 g moles of Na2SO4 = 4.32/142 = 0.03 Thus vol. = 0.03/.5 = 0.06 litres
Answered by ag_iitkgp - Thu Apr 26 03:51:03 2007

Hasselbach/pKa related questions.?
Q. Hi i'm stuck with the following questions: 3. Hydrochloric acid, HCl, is a strong acid. b) What pH value would a solution of 50 mmol/l HCl have? using the equation pH = -log10 [H+] 4. Acetic acid, CH3COOH, has a Ka value of 1.58 x 10-5 mol/l a) What does this mean? b) What is the pKa for acetic acid? c) Is acetic acid a weaker or a stronger acid than lactic acid (pKa = 3.7)? d) How would you prepare a 10 mmol/l solution, from a glacial acetic acid liquid acid solution, given that the molecular weight of acetic acid is 60 and its purity is 98% and its density (it is a liquid) is 1.05g/ml? I'm having trouble in calculating these. I'd appreciate if you know the answer, to include how you worked them out. thanks
Asked by 451317 - Sat Jan 26 15:39:05 2008 - - 1 Answers - 0 Comments

A. None of these involve Hasselbach. (3b) should come straight out of the definition of pH, since [H+] is equal to your HCl molarity (HCl being a strong acid). (4a): this is the definition of Ka; if you don't know it, it must be in your course materials. You really need to know it in order to move forward (4b) p(whatever) = - log (base 10) (whatever) (4c) so you can see that a higher pKa means a low Ka and a weaker acid (4d) This is just a sneaky bookkeeping exercise. 1) Work out how many g acetic acid you would need in a litre (multiply number of moles by molar mass) 2) Use density = mass/volume to find the volume containing that mass 3) (sneaky) DIVIDE by 0.98 because you need more than the answer from (2) because the acid is… [cont.]
Answered by Paul B - Sat Jan 26 19:35:26 2008

How do you calculate the concentrations of the conjugate acid/base forms in as initial solution?
Q. I have a biochemistry homework problem that I am stuck on. I. 0.0600 moles of NaOH is added to one liter of a 0.150 M glutamic acid solution initially at pH 4.100. (pk1=2.2, pk2= 4.2 and pk3=9.7). A. what are the concentration of the conjugate acid and base forms in the initial solution? B. What are the concentration forms in the final solution? C. what is the final pH of the solution? D. Would the final solution be better buffer against added or base? Why? E. What would the pH of one liter of water in which you placed 60.0 mmol of HCl? If you can help me with the hows to figuring these out it would be greatly appreciated.
Asked by Sheldon - Thu Sep 3 02:35:46 2009 - - 1 Answers - 0 Comments

A. First, make sure you understand the chemistry of the system: The pKa's are respectively associated with the following protons: COOH (main chain): pKa1 = 2.2 COOH (side chain): pKa2 = 4.2 NH (main chain):: pKa3 = 9.7 At pH = 4.1, the main chain carboxylate is nearly 100% deprotonated. (To make the chemistry clearer, at pH 3, say, glutamic acid is mostly present as a zwitterion with the main chain carboxylate mostly deprotonated and negatively charged, the side chain carboxylate still mostly protonated and neutral, and the -amino group entirely protonated and positively charged.) Anyway,... part A) at pH = 4.1, the side chain carboxylate is partially deprotonated: COOH + h o coo + H O ... Ka2 = [ COO ][H O ]/[ COOH] pKa2… [cont.]
Answered by Timothy - Sun Sep 6 14:34:35 2009

Organic chemistry calculation?
Q. how to calculate the mmol of 2mL HCl? of a concentrated HCl thank you
Asked by Leika M - Tue Jan 6 08:08:09 2009 - - 3 Answers - 0 Comments

A. Amount (moles) = concentration (M) x volume (L) or, if you prefer (as you should in this case) Amount (mmol) = conc (M) x vol (mL) So of course you need to know the conc of the HCl. That's all!
Answered by Paul B - Tue Jan 6 08:25:01 2009

Kinda hard titration question...?
Q. A mixed aqueous solution of HCl and maleic acid (HO2CCH=CHCO2H; Ka1 = 1.20 10-2; Ka2 = 5.37 10-7) is titrated with 0.1000-M NaOH. The first end point (congo red indicator) is reached after 42.42 mL of NaOH has been added. The second end point (cresol purple indicator) is reached after a total of 60.02 mL of NaOH has been added. What quantity of HCl (units of mol or mmol) was present in the initial mixture? This is due in an hour (11:30), it's my only problem left, and I can't figure it out. If anyone can lend any help, I would very much appreciate it. I'm desperate!
Asked by victoria - Sun Jul 15 22:15:53 2007 - - 3 Answers - 0 Comments

A. Tackle the 2nd endpoint 1st. 60.02 mL - 42.42 Ml = 17.6 M 17.6 mL x 0.1000 N = 1.76 meq of acid. The 1st endpoint 42.42 mL x 0.1000 N = 4.242 meq of acid 4.242-1.76 = 2.482 meq. HCl = 2.482 meq maleic acid = 1.76 millimoles (2x1.76 meq)
Answered by skipper - Sun Jul 15 23:16:14 2007

Help with Organic Chemistry?
Q. Have to come up with a procedures for "Borohydride Reduction of Vanillin to Vanillyl Alcohol". But have to start with 25.0 mmol of vanillin. I have googled for it, and most have much smaller starting amount for Vanillin (.4g). The over reaction is 4(C6H6O3) vanillin + 4H2O ---(with NaOH?)--> 4(Vanillyl alcohol) + H2BO3 + NaOH The ones i found on the website are: 1.) starting amout for vanillin=2.6 mmol 2.) 2.5 mmol of 1.0 M of NaOH which equals to 2.5 mL 3.) 2.1 mmol of Sodium BoroHydridewhich equals 80 mg 4.) less than 25 mmol of 2.5 M of HCl which equals 10 mL Can someone PLEASE show where these mmol came from through the equation just starting from 2.6 mmol of vanillin? The physical properties for Vanillin M.W. = 152.2, sodium… [cont.]
Asked by ==== - Thu Mar 13 22:47:47 2008 - - 2 Answers - 0 Comments

A. Ronald is correct, but I prefer a slightly different usage. For vanillin, 1 mole =152.2 g; 1 mmole = 152.2 mg 1 M = 1 mole/L = 1 mmol/mL 2.5 mL of 1 M solution = 2.5 mmol 2.1 mmol NaBH4 x 38 mg/mmol = 80 mg Incidentally, for 2.6 mmol of vanillin, 0.65 mmol NaBH4 theoretical; 0.65 mmol x 38 mg/mmol = 24.7 mg 80 mg is a pretty large excess. For common mg and mL lab use, mmol is convent as you can often do calculations without a calculator.
Answered by Dr OChem - Thu Mar 13 23:52:35 2008

Chemistry 11 Help Pleaseeee!!!!?
Q. 1.)Freons are compounds that are used as refrigerants. The molar mass of the freon with the formula CCl2F2(g), is Choose one answer. a. 120.91 g/mol b. 101.91 g/mol c. 66.46 g/mol d. 58.00 g/mol 2.)The amount of hydrochloric acid in 15 mL of 6.0 mol/L HCl(aq) is Choose one answer. a. 0.40 mmol. b. 2.5 mmol. c. 3.3 mmol. d. 90 mmol. 3.)How many grams of CuSO4 are required to mix 500. mL of 0.200 M solution? Choose one answer. a. 8.40 g b. 31.9 g c. 16.0 g d. 159.5 g 4.) The number of moles in 8.0 g NaOH is Choose one answer. a. 0.20 mol b. 0.33 mol c. 5.0 mol d. 320 mol 5.)The number of moles in 1.2 g of MgSO4 is Choose one answer. a. 0.010 mol b. 0.10 mol … [cont.]
Asked by Brittany - Thu Apr 23 18:34:06 2009 - - 1 Answers - 0 Comments

A. Visit these sites to learn more about chemistry formulas and concepts that can help you in regards to your problem:
Answered by Bawan - Mon Apr 27 17:04:17 2009

I need some help with a lot of grade 12 chemistry questions?
Q. 1) If 0.50 mol of iodine and 0.50 mol of chlorine are initially placed into a 2.00 L reaction vessel at 25 C, find the concentrations of all entities at equilibrium. I2(g) + Cl2(g) ---> 2ICl(g) K=81.9 at 25 C 2) Hydrogen chloride, HCl(g), decomposes into its elements according to the following equation: HCl(g) ---> H2(g) + Cl2(g) Teh equilibrium constant, K, is 3.2 * 10^-34 at 25 C. Calculate the equilibrium concentrations of all entities if 2.00 mol HCl(g) is initially placed in a closed 1.00 L vessel. 3) The following equation describes the formation of HI(g) H2(g) + I2(g) ---> 2HI(g) K=46.0 at 490 C Initially, 0.40 mol of hydrogen and 0.40 mol of iodine is injected into a 500 mL electrically heated vessel whose temperature is… [cont.]
Asked by cute_girly63 - Wed Jan 14 17:53:42 2009 - - 1 Answers - 0 Comments

A. You need to do your own homework, not waste time you could have used working on it, and asking other people to do it for you.
Answered by That one guy - Wed Jan 14 17:59:48 2009

PLEASE HELP ME! I NEED CHEMISTRY HELP?
Q. 1.3188 g of antacid is weighed and mixed with 75 mL of ecess 0.1746 M HCL. the excess acid required 27.30 mL of 0.09767 M NaOH for back titration. Calculate the neutralizing power of the antacid in terms of mmol H per gram of antacid. please help me
Asked by AriesPrincess~Slyffindor - Mon Sep 28 20:42:20 2009 - - 1 Answers - 0 Comments

A. This is book keeping. Find out how many mmol H+ you added as acid Subtract the amount that was left over for back titration The difference is the amount that was used up by the antacid Specifically,mmol H+ added = mL H+ x molarity H+ mmol left over = mmol OH- required to neutralise in back titration = mL OH- x molarity OH- Subtract the second number from the first, and that gives you mmol H+ removed by antacid Since you want mmol per g, just divide that last number by the mass (g) Then go back over the problem and make sure you understand why each step worked, and feeel the power! Good luck
Answered by Paul B - Tue Sep 29 09:07:48 2009

Consider the following solutions: CHEM II?
Q. Solution 1: 0.100 L of 0.25 M NaCH3CO2 + 0.050 L of 0.25 M HCl Solution 2: 0.100 L of 0.25 M HCH3CO2 + 0.050 L of 0.25 M NaOH a. What is the pH of this Solution 1? Ka for acetic acid is 1.8E-5. b. What is the pH of Solution 1 after adding 3.8 mmol of strong base? c. How many moles of acid does it take to change the pH of the original Solution 1 by 0.28 pH units? d. What is the pH of this Solution 2? Ka for acetic acid is 1.8E-5. e. What is the pH of Solution 1 after adding 3.8 mmol of strong base? f. How many moles of acid does it take to change the pH of the original Solution 1 by 0.28 pH units?
Asked by Anne C - Tue Apr 21 21:53:38 2009 - - 1 Answers - 0 Comments

A. Visit these sites to learn more about chemistry formulas and concepts that can help you in regards to your problem:
Answered by Bawan - Sat Apr 25 20:30:52 2009

A FEW BUFFER QUESTii0NS...?
Q. N0T 2 C0MPLiiCATED...JUST N0T sure where 2 START FR0M...Any help w0Uld be great...SH0W steps please && THANKS iiN advance 1. What is the pH of a solution prepared by dissolving 31 mmol of a weak base in 250 mL of a 0.139-M solution of its conjugate acid if pKa = 5.06 for the acid. 2. How many mL (to the nearest mL) of 0.182-M KF solution should be added to 360. mL of 0.189-M HF to prepare a pH = 2.60 solution? 3. How many grams (to the nearest 0.01 g) of NH4Cl (Mm = 53.49 g/mol) must be added to 700. mL of 1.294-M solution of NH3 in order to prepare a pH = 9.70 buffer? 4. What volume (to the nearest 0.1 mL) of 5.40-M NaOH must be added to 0.350 L of 0.350-M HNO2 to prepare a pH = 4.20 buffer? 5. What volume (to the nearest 0.1 mL) of… [cont.]
Asked by crazygirl0630 - Thu Nov 1 22:20:31 2007 - - 1 Answers - 0 Comments

A. 1. pH=pKa + log(nb/na) na= (x mL)(x M) =(250 mL)(0.139 M) =34.75 pH=5.06 + log(31/34.75) pH=5.01 hey, do you know this one: 1. Select all of the following that would be acceptable conjugate acid-base pairs to use in the preparation of a pH=9.8 buffer. HNO2/KNO2 HF/KF KHS/K2S KHSO3/K2SO3 HCN/KCN NH4Cl/NH3 KHCO3/K2CO3 H2S/KHS if you do, thanks!
Answered by fantastilistic3 - Fri Nov 2 03:00:11 2007