Number of results: 39,710
chem.
Thanks. I can help with this. And the experiment part of my question meant that if an experiment I would need to know what you were doing. As you have written it the problem now makes sense. 2NaOH + H2SO4 ==> Na2SO4 + 2H2O First, how many moles NaOH do you have in the ...
Saturday, March 19, 2011 at 4:10pm by DrBob222
chemistry
Do mean to calculate the volume of NaOH required for the neutralization? 2NaOH + H2SO4 ==> Na2SO4 + 2H2O moles H2SO4 = grams/molar mass Use the coefficients in the balanced equation to convert moles H2SO4 to moles NaOH. 15% NaOH means 15 g/100 mL (is that 15% w/v?) So ...
Thursday, March 17, 2011 at 4:30pm by DrBob222
Chemistry
25 mL 0.1 M acetic acid is titrated with 0.1 M NaOH. Calculate the pH after the addition of 0.0 mL, 1.0 ml, 2.5 mL, 4.0 mL, 7.0 mL, 10.5 mL, 15.0 mL, 17.0 mL, 20.0 mL, 22.0 mL, 23.0 mL, 23.5 mL, 24.0 mL, 24.25 mL, 24.5 mL, 24.75 mL, 25.0 mL, 25.25 mL, 25.5 mL, 25.75 mL, 26.0 ...
Monday, April 22, 2013 at 4:34am by smartkid <33
Chemistry
So far so good; however, you didn't include the volume of NaOH required in the titration. M(NaOH) x mL(NaOH) = M(vinegar) x mL(vinegar). 0.1250 x mL(NaOH) = M(vinegar) x 4.00 mL Plug in the mL NaOH, whatever it is, and you will have just one unknown which is the molarity ...
Wednesday, July 23, 2008 at 10:44pm by DrBob222
Chem
For the first one, check your work. I think you have an error of about 1000 times. I think 500*0.200/0.150 = 667 mL. However, that is the volume you will have, not how much water must be added. Water that must be added is 667-500 = ?? For the second one, M*mL = M*mL. 125 mL * ...
Friday, May 1, 2009 at 12:20am by DrBob222
Chemistry
M NaOH x mL NaOH = M HClO4 x mL HClO4 0.300M x mL NaOH = 0.15M x 30.0 mL You have one unknown; i.e., mL NaOH. Solve for that.
Wednesday, July 18, 2012 at 10:03pm by DrBob222
chemistry
500.0 mL x 0.200M = xmL x 15.0M(although you didn't write M). Solve for x and I get 6.667 mL. In practice you add 6.667 ml of the 15.0M NaOH to a 500.0 mL volumetric flask and add water to the mark on the flask. Note that you DO NOT add 6.667 mL NaOH to 493.3 mL H2O.
Sunday, November 6, 2011 at 7:43am by DrBob222
AP CHEMISTRY
mL NaOH x M NaOH = mL acid x M acid. mL NaOH x 0.125M = 25.0 x 0.175 Solve for mL NaOH. The 20% ionization is extraneous information.
Friday, March 29, 2013 at 1:20am by DrBob222
chemistry
The easy way is simple substitution. mL NaOH x M NaOH = mL HCl x M HCl mL NaOH x 0.15M = 35.00mL x 0.22M Solve for mL NaOH. I get (35.00*0.22/0.15) = approximately 51 mL NaOH. The only problem with working all titration problems that way is that not all titrations are 1:1 as ...
Friday, November 11, 2011 at 12:52am by DrBob222
Chemistry
You must know where the equivalence point is in order to know where you are on the titration curve. mL acid x M acid = mL base x M base. 25*0.434 = mL x 0.365 mL base = estimated 30 mL so 15 mL is BEFORE the eq point and 40 mL is AFTER the eq point. A. mols HI = M x L = ? mols...
Monday, June 3, 2013 at 10:03pm by DrBob222
chemistry
A volume of 100mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to the reaction flask. Classify the following conditions based on whether they are before the equivalence point, at the equivalence point, or after...
Monday, August 6, 2012 at 8:00pm by Anon
Chemistry - Molarity
I think something is missing but I don't know what. KHP + NaOH ==> H2O + NaKP H2P + 2NaOH ==> 2H2O + Na2P -------------------------- Suppose the sample is 100% KHP, then 25.0 mL x 0.127M = 3.175 mmols. That will take 3.175 mmols NaOH M NaOH = mmols/mL; ...
Monday, December 10, 2012 at 10:56pm by DrBob222
Chemistry
Include an explanation, balanced chemical equation, units, significant figures and/or calculations where appropriate. Standardization of NaOH Solution Trial 1: 0.501 g KHC8H4O4 15.22 mL NaOH Trial 2: 0.503 g KHC8H4O4 15.11 mL Trial 3: 0.500 g KHC8H4O4 15.10 mL Using the data ...
Monday, March 2, 2009 at 10:58pm by Jake
General Chemistry
"How would you prepare 150. mL of 0.1M sodium hydroxide given a stock solution of 3.0M NaOH?" I made calculations and figured the process would be to dilute 5 mL of the 3.0M NaOH to 150. mL. Is this correct? For the ideal 150. mL of 0.1M NaOH, I calculated there to ...
Sunday, April 21, 2013 at 9:18pm by Emily
Chemistry
I have interpreted this problem differently than Bob Pursley. I think the problem states that 85 mL HCl and 89 mL NaOH WERE LEFT when the accident was discovered; therefore, we have added 100-85 = 15 mL HCl and 100-89 = 11 mL NaOH. Therefore, mmoles HCl added = 15 mL x 0.06 M...
Thursday, July 9, 2009 at 3:16am by DrBob222
Chemistry
I can show you how to solve the problem but I have no numbers. However, mL vin x M vin = mL NaOH x M NaOH If you substitute mL vinegar and multiply by M vinegar (molarity = mols/L) and divide by M NaOH you can calculate mL NaOH. Most vinegars are about 5% acetic acid and I ...
Saturday, April 27, 2013 at 7:50pm by DrBob222
Chemistry
Calculate the pH of a 25 mL sample of distilled water after the addition of 1 mL, 2 mL, 3 mL, 4 mL, and 5 mL of NaOH and HCL. (0.1 M HCl and NaOH) In total, you need to show 10 calculations.
Monday, March 14, 2011 at 9:32pm by Mariah
college chem
Calculate the pH of a 25 mL sample of distilled water after the addition of 1 mL, 2 mL, 3 mL, 4 mL, and 5 mL of NaOH and HCL. (0.1 M HCl and NaOH) In total, you need to show 10 calculations.
Monday, March 14, 2011 at 9:50pm by Mariah
chemistry
What volume of WATER MUST BE ADDED to 10.0 mL of 6.0 M NaOH to make a solution that is 0.30 M in NaOH? Assume that the volumes are additive. a. 10 mL b. 190 mL c. 200 mL d. 210 mL e. 500 mL
Monday, March 29, 2010 at 6:30am by lawrence
Chemistry
A student begins with 25 mL of a 0.434 M solution of HI and slowly adds a solution of 0.365 M NaOH. 1. What is the pH after 15.00 ml of NaOH solution has been added? 2. What is the pH after 40 ml of NaOH solution has been added?
Monday, June 3, 2013 at 10:03pm by zac
chemistry
15.00 ml of NaOH is obtained in an Erlenmeyer flask. Before titrating with .2500 M HCl, the initial volume of acid in a buret is read as 2.08 ml. At the endpoint the final volume of acid in the buret is 25.20 ml. What was the concentration of NaOH?
Wednesday, November 28, 2012 at 7:32am by lisa
Chemistry
Molarity of NaOH solution = 0.06115 Volume of NaOH used = 15.50 mL Volume of unknown Sample = 10.00 mL Find the Moles of NaOH used? Find the molarity of H3PO4? Thanks!
Wednesday, October 10, 2012 at 8:18pm by Fred
chemistry
HC2H3O2 + NaOH ==> NaC2H3O2 + H2O initial: HC2H3O2 = M x mL = 0.1 x 8.00 mL = 0.8 mmoles. NaOH = M x mL = 0.1 x 12.90 mL = 1.29 mmoles. NaC2H3O2 = 0 final: You have more NaOH than HC2H3O2; therefore, all of the HC2H3O2 will be used and some of the NaOH will remain. ...
Monday, May 3, 2010 at 2:00pm by DrBob222
chemistry
Can you help me with the following: Calculate hte pH that results when 25 mL of .1M HCL is titrated with .1M NaOH solution run in from a buret at each of the stages of volume of .1 M NaOH add in 5.0 mL incrememnts beginning with 0 mL of NaOH
Friday, April 30, 2010 at 12:08pm by Cathy
chemistry
Hello, im having major issues with this problem...A 100.0-mL solution of 0.017 M CH3COOH (Ka = 1.8 10-5) is titrated with 0.025 M NaOH. Now, I've gotten some the answers all ready (right answers have *)....but three of them are wrong, can someone tell me what to do? 1.) ...
Wednesday, October 12, 2011 at 9:44pm by A person
college chemistry
You are given solutions of HCl and NaOH and must determine their concentrations. You use 38 mL of NaOH to titrate 100 mL of HCl and 23.2 mL of NaOH to titrate 50 mL of 0.0782 M H2SO4. Find the unknown concentrations.
Sunday, October 3, 2010 at 3:06pm by KELLY
Chemistry
I've been working on this problem for hours and cant figure it out. I know it should be simple, but I'm stuck... The question is... Describe how to prepare 1.00L of a 1.0% (w/v) NaOH from a 2.0 M NaOH. *I need to figure out how many ml of the stock solution is needed. ...
Thursday, January 10, 2013 at 6:08pm by James
chemistry
1). Adding NaOH to the buret after rinsing the buret with distilled water (essentially saying that you don't rinse the buret with NaOH)means that the NaOH is more dilute so it will take more NaOH from the buret to reach the equivalence point. So more mL means mL x M = a ...
Thursday, October 1, 2009 at 7:48pm by DrBob222
Chemistry
From what you've said, I think the teacher has TOLD you that the HCl s/he has provided is 0.1046 N. And if you've used 10.15 mL (0.01015 L), the # equivalents of HCl is L x N = 0.1046 x 0.01015 = ? Next, I assume you are to calculate the N of the NaOH listed next at 10...
Saturday, March 3, 2012 at 9:35am by DrBob222
Chemistry-Is this correct?
The concentration of NaOH is normally in the range of 3-6 M when they peel potatoes and soak them in a solution of NaOH , remove them and spray them off. If I have 45.7 mL of 0.500M of H2SO4 required to react completely with a 20.0 mL sample of NaOh. What is the molar ...
Thursday, May 20, 2010 at 4:20pm by Sara
Chemistry
moles H2X present = 0.1873/85.00 = ?? moles NaOH = 2 x that number. MNaOH = moles NaOH/L NaOH Solve for L NaOH and convert to mL. 41.89 mL is the correct answer.
Wednesday, May 11, 2011 at 9:06pm by DrBob222
Chemistry
The density of the solution is 1.17 g/mL. How much does a liter weigh? 1.17 g/mL x 1000 mL = ?? grams = mass of soln. How much of that is NaOH? 0.0357%; therefore, ?? g soln x 0.0357 = ??g NaOH.(The remainder is the mass of the water in the solution). How many mols NaOH is ...
Sunday, September 7, 2008 at 1:24pm by DrBob222
chemistry
11.During an acid-base titration, 25 mL of NaOH 0.2 M were required to neutralize 20 mL of HCl. Calculate the pH of the solution for each of the following: 12.Before the titration. 13.After adding 24.9 mL of NaOH. 14.At the equivalence point. 15.After adding 25.1 mL of NaOH. ...
Wednesday, May 8, 2013 at 9:18pm by Tina
Chemistry
I'm a little confused as to the question. If you want percent water (not counting the solutions) it is (4/10)*100 = ? If you want to count the water in the aq NaOH too that is almost 2 mL (0.01M NaOH is only 0.4 g NaOH in a liter which will be 0.4 mg in 1 mL or 0.8 mg in 2...
Thursday, February 9, 2012 at 9:45pm by DrBob222
chemistry
Hello, im having major issues with this problem...I need to find a pH for a 100.0-mL solution of 0.017 M CH3COOH (Ka = 1.8 10-5) is titrated with 0.025 M NaOH. Now, I've gotten some the answers all ready (right answers have *)....but three of them are wrong, can someone ...
Wednesday, October 12, 2011 at 11:20pm by A person
Chemistry
Which of the following solution should be mixed with 50.0 mL of 0.050 M HF solution to make an effective buffer? A) 50.0 mL of 0.10 M NaOH B) 25.0 mL of 0.10 M NaOH C) 50.0 mL of 0.050 M NaOH D) 25.0 mL of 0.050 M NaOH
Sunday, April 1, 2012 at 9:20am by Anon
Chemistry
c1v1 = c2v2 c = concn v = volume For example. You have 2M NaOH and you want to make 500 mL of 0.25M. How much of the 2M stuff do you take? 2M*x mL = 0.25M*500 mL mL = (0.25*500/2) = 62.5 mL So you take 62.5 mL of 2M NaOH, add to 500 mL volumetric flask and make to the mark. ...
Wednesday, January 16, 2013 at 10:41pm by DrBob222
Chemistry
A 50.0 mL sample of 0.12 M formic acid, HCOOH, a weak monoprotic acid, is titrated with 0.12 M NaOH. Calculate the pH at the following points in the titration. Ka of HCOOH = 1.8 multiplied by 10-4. What is the pH when 50.0 mL NaOH is added 60.0 mL NaOH is added 70.0 mL NaOH is...
Sunday, April 3, 2011 at 5:13pm by George
Chemistry
Look at titrating the first H, then double it. H3PO4 + NaOH ==> NaH2PO4 + H2O mols H3PO4 = 25 mmols. So you will need 25 mmols NaOH to react with the first H of H3PO4. M NaOH = mols NaOH/L NaOH 2.0M = 25 mmols/mL mL = 25/2.0 = 12.5 mL. So it will take 25 to neutralize ...
Saturday, March 23, 2013 at 5:32pm by DrBob222
Chemistry
HCl + NaOH ==> NaCl + H2O How many milliequivalents(m.e.) HCl were used? m.e. = mL x N A m.e. of HCl = m.e. NaOH = m.e. of anything. Then N = m.e. NaOH/mL NaOH.
Saturday, June 25, 2011 at 4:36am by DrBob222
Chemistry
One commercial method used to peel potatoes is to soak them in a solution of NaOH for a short time, remove them from the NaOH, and spray off the peel. The concentration of NaOH is normally in the range of 3 to 6 M. The NaOH is analyzed periodically. In one such analysis, 45.7 ...
Wednesday, January 12, 2011 at 1:06am by Chin
chemistry, please help!
Given: 1. Mass of oxalic acid + weighing paper 2.0355g 2. Mass of weighing paper 0.5219 g 3.volume of oxalic acid solution 250 mL 5. Volume of oxalic acid used in titration trial 1: 15.o mL Trial 2: 15.0 mL 6. Volume of NaOH solution used in titration: Trial 1: 12.1 mL trial 2...
Monday, May 2, 2011 at 12:46am by Tina
Chemistry
Titration of 25.00 mL of a monoprotic acid solution with 0.2269 M NaOH is: Initial NaOH = 3.96 mL Final NaOH = 38.84 mL (dif is 34.88mL) Based on data what is the molar concentration of acid? I did V x M = 0.03488L NaOH x 0.2269 = 0.079143 Not sure if I should have switched to...
Friday, December 7, 2012 at 11:44am by Terry
Chemistry help
There are numerous ways to do this but the shortest way is to use the dilution formula that you have in the last sentence of your post. I even take a short cut with that. 0.01M NaOH x (2.0 mL/8 mL) = ? M. Note: It isn't needed, but you really do have the mass. moles NaOH...
Thursday, March 15, 2012 at 5:05pm by DrBob222
Chemistry
An unknown amount of water is mixed with 350 mL of a 6 M solution of NaOH solution. A 75 mL sample of the resulting solution is titrated to neutrality with 51.2 mL of 6 M HCl. Calculate the concentration of the diluted NaOH solution. Answer in units of M What volume of water ...
Monday, May 13, 2013 at 6:07pm by Anonymous
Chemistry
An unknown amount of water is mixed with 310 mL of a 6 M solution of NaOH solution. A 75 mL sample of the resulting solution is titrated to neutrality with 58.2 mL of 6 M HCl. Calculate the concentration of the diluted NaOH solution. Answer in units of M What volume of water ...
Monday, March 19, 2012 at 12:28am by ag
Chemistry
An unknown amount of water is mixed with 310 mL of a 6 M solution of NaOH solution. A 75 mL sample of the resulting solution is titrated to neutrality with 58.2 mL of 6 M HCl. Calculate the concentration of the diluted NaOH solution. Answer in units of M What volume of water ...
Monday, March 19, 2012 at 12:27am by ag
Chem
Vinegar (you can look on the bottle to see) usually is about 5% acetic acid. Convert that 5% (some are 4%) to molarity, then start with enough vinegar to use approximately 40 mL of the NaOH. You can use mL vin x M vin = mL NaOH x M NaOH. Will this get you started?
Sunday, October 10, 2010 at 1:54am by DrBob222
Chemistry
You weighed KHP. Calculate the moles KHP you titrated with NaOH. Moles KHP = grams/molar mass. You should write the equation for KHP + NaOH but it is a 1:1 ratio (1 mole KHP to 1 mole NaOH). Therefore, moles KHP titrated will equal moles NaOH used. Then you know M x L = moles ...
Sunday, September 20, 2009 at 2:05pm by DrBob222
chemistry
(1) mols KHC8H4O4 = grams/molar mass (2)mols NaOH = mols KHC8H4O4 (since the mols ratio is 1:1 in the titration. Write and balance the equation to confirm that.) (3) M NaOH = mols NaOH/L NaOH. So you over titrate with the base. No mL are involved in equation 1 above. No change...
Tuesday, May 7, 2013 at 3:27pm by DrBob222
Chemistry
5.88% v/v means 5.88 mL acetic acid/100 mL solution. m = v*d; therefore 5.88 mL x 1.409 g/mL = grams acetic acid. Convert grams to moles. moles = grams/molar mass. Convert to M = moles/L of soln. Then write the equation. CH3COOH + NaOH ==> CH3COONa + H2O moles CH3COOH ...
Sunday, January 16, 2011 at 9:54pm by DrBob222
chemistry
how many mL of 0.15 M NaOH solution are required to neutralize 35.00 mL of 0.22 M HCl
Friday, November 11, 2011 at 12:52am by sonya
chem
how many ml of 0.15 m naoh solution are required to neutralize 35.00 ml of 0.22 m hcl
Friday, April 23, 2010 at 12:24am by Anonymous
chemistry
20 ml of 0.1 M Acetic acid and 25 ml of 0.1 M Sodium Acetate are mixed together. then 5ml of 0.1 M of NaOH is added. Find the Ph before the addition of NaOH and after the addition of NaOH??
Tuesday, March 15, 2011 at 12:54am by james bond
chem
20 ml of 0.1 M Acetic acid and 25 ml of 0.1 M Sodium Acetate are mixed together. then 5ml of 0.1 M of NaOH is added. Find the Ph before the addition of NaOH and after the addition of NaOH??
Tuesday, March 15, 2011 at 12:41am by james bond
Chemistry
The easy way to do this is mL NaOH x M NaOH = mL HCl x M HCl This equation works only when the reactants are 1:1 in the equation as is the case with HCl and NaOH. NaOH + HCl ==> NaCl + H2O
Thursday, June 7, 2012 at 11:31am by DrBob222
chemistry
Calculate the pH of a solution obtained by mixing 30 mL of 0.75 M CH3COOH with 15 mL of 1.5 M NaOH.
Tuesday, January 24, 2012 at 9:58pm by grewal
chemistry
how many ml of 0.15 NaOH solution are required to neutralize 35.00 ml of 0.22 moles of HCL
Wednesday, April 14, 2010 at 1:43pm by mc
Chemistry
Include an explanation, balanced chemical equation, units, significant figures and/or calculations where appropriate. Determination of the Molar Mass of an Unknown Acid Trial 1: 0.499 g unknown acid 24.07 mL NaOH Trial 2: 0.501 g unknown acid 23.71 mL NaOH Trial 3: 0.500 g ...
Monday, March 2, 2009 at 11:00pm by Jake
Chem
25.0 mL of 0.100 M acetic acid (Ka= 1.8 x 10^-5) is titrated with 0.100 M NaOH. Calculate the pH after the addtion of 27.00 mL of 0.100M NaOH. my work CH3COOH + H2O <-> H3O^+ + CH3COO^- 25mL x 0.100 mmol/ml = 2.5mmol CH3COOH 27mL x 0.100 mmol/ml = 2.7mmol NaOH ...
Tuesday, June 29, 2010 at 4:02pm by Anonymous
Chem 110
Start with a balanced equation HCl + NaOH -> NaCl so one mole of HCl reacts with 1 mole of NaOH Thus we need to find the volume of HCl that contains 0.15 mole of HCl We can do this my simple proportion 1 litre (or 1000 ml) of HCl contains 2.0 moles 1 ml of HCl containes...
Monday, November 8, 2010 at 5:13am by Dr Russ
Chemistry
How many moles of acetic acid do you have. M x L = ?? How many moles NaOH will that require? It must be the same because the reaction between acetic acid and NaOH is 1:1. Then moles NaOH = M x L; you have moles NaOH and M NaOH, solve for L. For 1:1 reactions, the shorter mL x ...
Saturday, November 27, 2010 at 5:44pm by DrBob222
Chemistry
We've been here before. What's the problem? You have M NaOH (from the pH = 13.68), you have mL HClO4 and you have M HClO4. There is only one unknown in mL NaOH x M NaOH = mL HClO4 x M HClO4.
Thursday, March 8, 2012 at 11:50pm by DrBob222
College Chemistry
HCl + NaOH ==> NaCl + H2O Since this is a 1:1 reaction, one may use this formula. mL x M = mL x M mL NaOH = 39.39 - 2.44
Sunday, October 3, 2010 at 3:14pm by DrBob222
Chemistry
14-pH=pOH=14-10.5=3.5 pOH=-log[OH-]=3.5 10^-(pOH)=10^-(3.5)=OH concentration in M 4.40L*(OH concentration in M)= moles of OH moles of OH*(39.997 g/mol)= mass of NaOH needed mass of NaOH needed Since 15% is 0.15g of NaOH/g of solution, solving for g of solution mass of NaOH ...
Sunday, March 17, 2013 at 1:58pm by Devron
chemistry
Draw the tritration curve for the neutralization of a 100ml sample of 1M solution of HCl with a .6M solution of Naoh point 1: 2 ml NaOH edded point 2: 40 ml NoOH ADDED POINT 3: EQUIVALENCE POINT 4: 80 ML NaOH added
Sunday, April 3, 2011 at 1:55pm by sam
Chemistry
One example each for 0.1M HCl/NaOH added to 25 mL H2O. The diluted HCl solution is 0.1M x (1 mL/26 mL) = (HCl) pH = -log(HCl) The diluted NaOH solution is 0.1M x (1 mL/26 mL) = (NaOH) pOH = -log(NaOH). Then pH + pOH = pKw = 14.00
Monday, March 14, 2011 at 9:32pm by DrBob222
Chemistry!!!!!! Please help immediately!!!
Please show me step by step. Thank you. One commercial method used to peel potatoes is to soak them in a solution of NaOH for a short time, remove them from the NaOH, and spray off the peel. The concentration of NaOH is normally in the range of 3 to 6 M. The NaOH is analyzed ...
Sunday, March 24, 2013 at 6:21pm by Lilyyyyyyyyyyyy!!!!
Chemistry
Запишем уравнение реакциÐ...
Tuesday, April 13, 2010 at 8:44am by Yevgeniy
chemistry
Consider the reaction HCl + NaOH ->NaCl + H2O Given: HCl Solution: 22 degrees celsius NaOH Solution: 22 degrees celsius Final Temperature: 26.1 degrees celsius A. Calculate the amount of heat evolved when 15 mL of 1.0 M HCl was mixed with 35 mL of 1.0 M NaOH B. ...
Monday, March 28, 2011 at 12:11pm by Vanessa
chem-acid base titrations
HCl + NaOH ==> NaCl + H2O mols HCl used = L x M = 0.0200 L x 0.2019 M = ?? mols NaOH (from the equation it is a 1:1 ratio of HCl to NaOH; therefore, mols NaOH must be the same as that for HCl). mols NaOH = L x M You have L (0.03963 mL) and you have mols (from HCl), ...
Monday, April 7, 2008 at 8:12pm by DrBob222
Chemistry
Usually you have mL NaOH but you also know how many mL of the acid and the molarity of the acid. With no more information than mL NaOH, no, it can't be done.
Wednesday, May 12, 2010 at 11:48pm by DrBob222
Organic Chemistry
yes. that would be 4.1 mL of 3 M NaOH to add, theoretically, and if that is 3 M and not 3.0 M, then you may have only one s.f. and your 4 would be right (but not 4.0). If it is 3.0 M NaOH, I would round to 4.1 mL. Since benzoic acid can be titrated with NaOH, that really isn&#...
Thursday, January 20, 2011 at 10:59pm by DrBob222
Chemistry
What volume of a 15.0% by mass NaOH solution, which has a density of 1.116 g/mL, should be used to make 4.40 L of an NaOH solution with a pH of 10.5?
Sunday, March 17, 2013 at 1:58pm by Emma
chemistry
how do you solve this What is the pH of the solution created by combining 2.60 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)? mL NaOH 2.60 pH wHCl: _ pH wHC2H3O2 : _ Complete the table below: What are the pH values if you ...
Monday, July 12, 2010 at 6:03pm by Jin
chemistry
mols HCl = 0.0920 x 0.025 = 0.0023 mols NaOH = 0.100 x 0.00015 = 0.15E-5 mols HCl remaining = 0.0023-1E-5 = 0.002285 MHCl remaining = 0.002285 mols/(0.025+0.00015)L = 0.09085 pH = -log(0.09085) = about 1.04 for 0.15 mL NaOH. Devron is correct for 23.0 mL 0.1M NaOH; the pH = 7....
Wednesday, January 23, 2013 at 12:08pm by DrBob222
chemistry
A sample of carboxylic acid is added to 50.0 mL of 2.27 M NaOH and the excess NaOH required 28.7 mL of 1.86 M HCl for neutralization. How many moles of NaOH reacted with the carboxylic acid?
Thursday, November 11, 2010 at 5:43pm by Ali
biochemistry
A). pH = -log(H^+) 7.6 = -log(H^+). Solve for (H^+). You should get approximately 2.5E-8. B) 150 mL x 0.1M NaOH = 15 millimiles. 200 mL x 0.1M benzoic acid = 20 mmols. Let's call benzoic acid HB, then ............HB + NaOH ==> NaB + H2O initial....20......0........0...
Friday, September 21, 2012 at 9:38am by DrBob222
chem lab (webwork)
What is the pH of the solution created by combining 1.80 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)? mL NaOH pH wHCl pH wHC2H3O2 1.80 Complete the table below: What are the pH values if you take into account that the 8....
Thursday, April 15, 2010 at 11:05pm by anonymous
chemistry
mols HCl = M x L. M you have as 0.2500. L. Final 25.20mL - 2.08 mL = mL HCl. Convert to L and find mols as above. Then mols NaOH = mols HCl. M NaOH = mols NaOH/L NaOH. Solve for M.
Wednesday, November 28, 2012 at 7:32am by DrBob222
Chemistry
Which one of the following pairs of 0.100 mol L-1 solutions, when mixed, will produce a buffer solution? A. 50. mL of aqueous CH3COOH and 25. mL of aqueous CH3COONa B. 50. mL of aqueous CH3COOH and 25. mL of aqueous HCl C. 50. mL of aqueous NaOH and 25. mL of aqueous HCl D. 50...
Sunday, May 2, 2010 at 11:27pm by Macky
Chemistry
moles chloroacetic acid = M x L moles acid = moles NaOH M NaOH = moles NaOH/L NaOH. You have M NaOH and M NaOH, solve for L NaOH and convert to mL if you wish.
Thursday, January 19, 2012 at 9:25am by DrBob222
chemistry
A 20.1 mL sample of a weak monoprotic acid, HX, requires 50.0 mL of 0.060 M NaOH to reach the equivalence point. After the addition of 30.0 mL of NaOH, the pH is 4.90. What is the Ka of HX?
Sunday, March 27, 2011 at 10:46pm by Anonymous
Chemistry
It depends upon which acid is being titrated with NaOH because the equivalence point is NOT always at 7.0. However, again, I assume this is a simple case. Use mL(acid) x N(acid) = mL(NaOH) x N(NaOH). The only unknown is N(acid).
Wednesday, February 17, 2010 at 1:10pm by DrBob222
chemistry
How many moles NaOH do you need? THat is M x L = ? M NaOH = moles NaOH/L NaOH. YOu know M NaOH (6.0M) and you know moles from the first part. Solve for L and convert to mL.
Monday, October 24, 2011 at 7:10pm by DrBob222
Chemsitry
Hi I have done this problem like 15 times when I find the mole I get .8 for HCL and .23 for Naoh and the equilibrium conc i get is negative. then I plug it in and get it wrong. Can someone please help me? Complete the table below: Note: Make simplifying assumptions, do not use...
Friday, April 27, 2012 at 9:07pm by someone
chem 2
which one of the following pairs of 0.100 mol L-1 solutions, when mixed will produce a buffer solution? a 50. mL of aquesous CH3COOH and 25.mL of aquesous HCI b 50. mL of quesous CH3COOH and 100. mL of aqueous NaOH c 50. mL of aquesous NaOH and 25. mL of aqueous HCI d 50. mL ...
Saturday, March 3, 2012 at 1:42pm by leaha
Chemistry
A complicated work but do-able. NaOH + HCl ==> NaCl + H2O. I think one must decide where we are at the equivalence point. Since the M NaOH is twice that of the HCl, that means the HCl will take twice as much to neutralize the NaOH. So that 35 mL is made up of 2 mL HCl ...
Sunday, May 2, 2010 at 11:52pm by DrBob222
chem SCC 151
My crystal ball isn't working well today. You don't have nearly enough information to answer a question. You need volume NaOH. Use this mL HCl x M HCl = mL NaOH x M NaOH and solve for the M HCl. Post your work if you get stuck.
Sunday, July 15, 2012 at 2:56pm by DrBob222
Chemistry
I don't know how you came up with those numbers. 2KOH + H2SO4 ==> K2SO4 + 2H2O moles H2SO4 = M x L = ?? Use the coefficients to change moles H2SO4 to mole KOH. You can see that moles NaOH = 2x moles H2SO4 Then MNaOH = moles NaOH/L NaOH Then convert L NaOH to mL. You...
Friday, March 18, 2011 at 11:40am by DrBob222
chemistry
15.0 mL of 8.00 M NaOH was diluted with water to 500 mL. Calculate the molarity of the diluted NaOH solution. My calculations: 0.015L x 8.00moles/1L x 39.997/1mole = 4.8g M= (4.8g x 1mole/39.997) / 0.500L = 0.24M is my workings correct? Is there another way to get to this?
Monday, April 22, 2013 at 5:55am by alex
chemistry
The term 1% NaOH (m/v) means 1 g NaOH in 100 mL solution. 1g/100 mL then gives you the g/mL and that x 5 gives you the g in 5 mL.
Friday, April 25, 2008 at 1:31pm by DrBob222
chemistry
How many milliliters of 0.729 M NaOH are needed to prepare 150.0 mL of 0.122 M NaOH? 25.1 mL
Wednesday, February 22, 2012 at 6:34pm by krystal
chemistry
HCl + NaOH ==> NaCl + H2O At zero mL. YOu have 0.1 M HCl so the pH is pH = -log(H^+). Since HCl is a strong acid, H^+ = 0.1 M and pH = 1. All of the others are done this way: a. moles HCl initially = M x L = 0.025 x 0.1 = 0.0025 moles HCl. b. moles NaOH added. (at 5 mL ...
Friday, April 30, 2010 at 12:08pm by DrBob222
Chemistry
when 10.0 ml of a .20 M Cu (NO3)2 solution is mixed with 15.0 mL of 0.20M NaOH, what is the theoretical yield (grams) of copper (II) hydroxide?
Wednesday, September 12, 2012 at 7:17pm by Lily
science
You add 12.5 mL of 4.15 acetic acid to 25.0 mL of 1.00 M NaOH. Calculate the hydronium ion concentration and pH of the resulting solution.
Monday, March 7, 2011 at 5:45pm by Bill
Chemistry lab
NaOH + H3PO4 ==> NaH2PO4 + H2O mL x M NaOH = millimols. That = mmols H3PO4 M H3PO4 = mmols H3PO4/mL H3PO4 = ? 2NaOH + H3PO4 ==>Na2HPO4 mL x M NaOH = 14.861 x 0.1538 = ? mmols H3PO4 = ? x 1/2 M H3PO4 = mmols H3PO4/10 mL = ?
Wednesday, April 11, 2012 at 11:16am by DrBob222
Chemistry II
A 12.00 mL sample of sulfuric acid from an automobile battery requires 34.62 mL of 2.42 M sodium hydroxide solution for complete neutralization. What is the molarity of the sulfuric acid? H2SO4 + 2NaOH ==> Na2SO4 + 2H2O molarity NaOH x liters NaOH = mols NaOH. mols ...
Wednesday, April 25, 2007 at 6:33pm by Jayd
CHEMISTRY (WEBWORK)
What is the pH of the solution created by combining 1.20 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)? mL NaOH pH w/ HCl pH w/ HC2H3O2 1.20 ? ? Complete the table below: What are the pH values if you take into account ...
Monday, July 12, 2010 at 6:19pm by Anonymous
Chemistry
How many ml of .050 M NaOH are required to prepare 1500.0 mL of .0020 M NaOH? Please show work, I'm not sure how to do this. Thanks!
Tuesday, September 1, 2009 at 9:48pm by Candace
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