April 21, 2014

Search: NaOH

Number of results: 4,667

If molecule of NaCl is heavier than a molecule of NaOH and NaOH is heavier than a molecule of HCl, can you explain the different conductivites you observed for these compounds?
Friday, October 2, 2009 at 7:33am by roek

from the abbreviated formula it is a monoprotic acid. Thus the equation for the neutralisation is HGly + NaOH -> H2O + NaGly so one mole of HGly reacts with one mole of NaOH. number of moles of NaOH used is 0.02750 L x 0.120 mole L^-1 which is the same as the number of ...
Tuesday, December 14, 2010 at 8:36am by Dr Russ

HCl is a strong acid and ionizes 100%. NaOH is a strong base and ionizes 100%. We have no way of knowing the pH of HCl + H2O because the problem doesn't state the amount of HCl added or the concentration of HCl. For the NaOH + HCl ==> NaCl + HOH. NaCl in water is a neutral ...
Tuesday, May 15, 2007 at 5:10pm by DrBob222

You have two questions here. #1. pH of 0.8 mL x 0.1 M NaOH + 8.00 mL of 0.1 M HCl. moles NaOH = M x L = 0.1 x 0.0008 = 0.00008 moles NaOH. moles HCl = M x L = 0.1 x 0.008 = 0.0008 moles NaOH + HCl ==> NaCl + H2O Now place the moles under the reactants so you can see what is...
Monday, July 12, 2010 at 4:33am by DrBob222

mols HCl initially = 0.025 x 0.1 = 0.0025 To end up with pH = 3 (0.001M H^+) we want H^+ to be 1E-3. If we work in millimols and let x = mL of 0.1M NaOH, we have then [(25.00 x 0.1M)-(0.1x)/(25+x)] = 1E-3M Solve for x and I obtained approx 24.5 mL of 0.1M NaOH that must be ...
Wednesday, February 26, 2014 at 7:44pm by DrBob222

I there, this is my lab worksheet, I am having trouble filling it out. Please help me with it. Expt #1- Molecular Weight of Unknown Acid Unknown Acid: #3 Mass of Unknown solid acid transferred:1.0g Volume of volumetric flask: 100.00 mL Concentration of NaOH: 0.0989 M Aliqot of...
Monday, October 4, 2010 at 10:17pm by Steve

On the axis below, draw the titration curve for the titration of CH3COOH (a weak acid) with NaOH (a strong base). In the titration, the equivalence point occurs when 40 mL of NaOH has been added. Make sure to label your x- and y-axis and put numbers on your grid. Make sure ...
Monday, April 23, 2012 at 12:36am by Anonymous

mols HCl = M x L = ? mols NaOH = mols HCl Then M NaOH = mols NaOH/L NaOH
Thursday, February 20, 2014 at 8:47pm by DrBob222

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

Write the equation. HCl + NaOH ==> NaCl + H2O mols NaOH = M*L = 0.10 M x 0.1 L = ?? mols HCl = M*L = 0.15 M x 0.075 L = ?? See which is in excess, how much is left over after reacting, then pH = - log(H^+). Post your work if you get stuck.
Wednesday, August 22, 2007 at 10:10pm by DrBob222

This may help. ............NaOH + HCl ==> NaCl + H2O initial....0.001..0.009.....0......0 change...-0.001..-0.001....+0.001..0.001 equil.......0.....0.008..... The base and acid react. So 0.001 moles NaOH "disappear" and take 0.001 moles HCl with it (to make 0....
Tuesday, November 8, 2011 at 7:12pm by DrBob222

More Solutions Chem
first we balance the given chemical equation, thus: 2NaOH + H2SO4 -> Na2SO4 + 2H2O note that we need to balance a chemical equation because we need their stoichiometric coefficients (number before the chemical formula of a compound or element) to make ratios. since we are ...
Friday, April 15, 2011 at 11:42am by Jai

CHEMISTRY (Help please)
Compare the results of the two measurements of a heat of neutralization (HCl&NaOH=11.58 kcal/mol) (CH3COOH&NaOH=13.06kcal/mol). Estimate the error in this. How close would they have to be to be considered identical? Quote numbers to justify your answer
Thursday, April 8, 2010 at 1:51am by Michelle

If you have 20.0g of a Maalox tablet, and 40mL of .1 M HCI, how much extra help from NaOH would you need to bring the acid to neutral? Back titrate using .05 mL NaOH, drop by drop. Please put this into a formula for me to figure out. Thanks.
Tuesday, March 26, 2013 at 7:30pm by Pam

When the concentration of CH3Br and NaOH are both 0.150M, the rate of the reaction is 0.0020M/S. (a) What is the rate of the reaction if the concetration of CH3Br is doubled? _______________M/s (b) What is the rate of the reaction if the concentration of NaOH is halved? ...
Wednesday, February 27, 2013 at 1:49pm by dj

Write the equation and balance it. Then determine the moles of NaOH and moles acetic acid. M x L = moles. Next you must look at the moles and determine which is in excess, if either, and recognize the kind of solution produced. If an excess of NaOH, the pH will be determined ...
Saturday, April 24, 2010 at 10:41pm by DrBob222

Convert 6.41g NaOH to moles. moles = grams/molar mass. Call this soln 1. Convert the 0.25O M solution to moles. moles = M x L = ? Call this soln 2. Convert 2.28 gallons H2O to L. Add moles soln 1 to moles soln 2. Then total moles NaOH/L (from the 2.28 gallons) = M of the final...
Tuesday, March 22, 2011 at 8:25pm by DrBob222

While titrating 25.00mL of a weak acid, HA, with 0.1500M NaOH, you reach equivalence point after adding 27.00mL of the NaOH. The pH of the acid initially was 2.48. What is the dissociation constant of the acid?
Monday, October 27, 2008 at 9:38pm by Sam

Given: H3PO4(aq) + 3 NaOH(aq) yields 3 H2O(l) + Na3PO4(aq) change in H = -166.5 kJ What is the value for q (heat) if 4.00 g of NaOH reacts with an excess of H3PO4? Please can you explain to me HOW to do this, not just give me an answer? Thanks!
Sunday, November 30, 2008 at 1:20pm by Jake

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

A 50mL solution of Histidine-HCl (Histidine: pKa1=1.8, pKa2=9.2, pKa3=6.0)is titrated with 0.500M NaOH. What is the expected pH of the histidine solution at the points in the titration when 14.0mL, 26mL, and 38mL of the NaOH titrant is added?
Wednesday, October 2, 2013 at 10:53pm by Drew

You have a vinegar solution you believe to be 0.8 M. You are going to titrate 21.39 mL of it with a NaOH solution that you know to be 0.774 M. At what volume of added NaOH solution would you expect to see an end point? Answer in units of mL
Tuesday, March 11, 2014 at 9:07pm by Sara

11th grade Chemistry
[mole-mole] How many moles of NaOH are needed to react with 5.0 moles of sulfuric acid, H2SO4??? Equation: H2SO4 + 2 NaOH ---> Na2 SO4 +2 H2 O
Monday, March 30, 2009 at 7:39pm by Jenny

4.00 grams of NaOH have been added to 2.0 liters of water. The molarity of this solution is ____M. the concentration of this solution is ______ grams of NaOH per liter of water
Friday, August 6, 2010 at 10:50am by Maria

what is a titration of 25 ml of 0.15 M HCLO2 is performed with 0.15 M NaOH. What is the pH of the solution after 25 ml of naOH have been titrated? Ka = 0.011
Monday, August 12, 2013 at 5:46pm by auriane

Naoh + CH3COOH = CH3COONa + H2O So i am going titrate acid(CH3COOH) to base (Naoh). How do i measure solubility of Acetic Acid( CH3COONa ) from that? check when it forms a precipitate?
Monday, June 2, 2008 at 11:34pm by chemistrybuddy

Chemistry 11
the concentration of HCl, a strong acid is 0.500 M. if 20.0 ml HCl is needed to titrate 40.0 ml of NaOH, what is the concentration of NaOH Help me please. I'm confused by this
Thursday, April 26, 2012 at 3:40am by Jessie

knowing that one mole of KHP, C8H5O4K, reacts with one mole of NaOH, what mass of KHP is required to neutralized 30.0mL of the 0.10 M NaOH solution?
Tuesday, October 23, 2012 at 3:44pm by tierra

chem-acid-base titrations
1/ HC2H3O2 + NaOH ==> HOH + NaC2H3O2 2. Calculate mols NaOH from L x M = ?? Convert mols NaOH to mols HC2H3O2 using the coefficients in the balanced equation. M acetic acid= mols/L. 3. Determine mols acetic acid in the 10 mL sample and from that grams acetic acid, then ...
Monday, April 21, 2008 at 10:06pm by DrBob222

Lets say that the NaOH solution is X M. Number of millimoles of NAOH used is 24.25 x X each millimole of NaOH neutralises half a millimole of H2SO4 so number of millimoles of H2SO4 is 24.25 x X x 0.5 if C (in mole per litre) is the concentration of H2SO4 then 26.00 x C = 24.24...
Monday, November 3, 2008 at 10:20am by Dr Russ

1) Write a balanced chemical equation for the reaction between Cu(NO3)2 * 3 H2O and NaOH. Underline the formula for the precipitate produced by this reaction. (The water of hydration in Cu(NO3)2 * 3 H2O appears as liquid water on the right side of the equation) 2) Calculate ...
Monday, May 17, 2010 at 6:59pm by Sarah

A 30.0-ml volume of 0.50 M CH3COOH (Ka=1.8*10^-5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH.
Monday, June 21, 2010 at 9:07am by Syra

A volume of 25.0 ml of 0.100 M CH3CO2H is tiltrated with 0.100 M NaOH. What is pH after the addition of 12.5 mL of NaOH? (Ka for CH3CO2H = (1.8 x 10-5)
Wednesday, October 20, 2010 at 12:22am by Helen

The beaker contains 0.3000 M HCl and the buret contains 0.3000 M NaOH. Write a complete balanced equation for the neutralization reaction between HCl and NaOH.
Sunday, March 27, 2011 at 1:19pm by Kat

A volume of 25.0 mL of 0.100 M CH3CO2H is titrated with 0.100 M NaOH. What is the pH after the addition of 12.5 mL of NaOH? (Ka for CH3CO2H = 1.8 x 10-5)
Tuesday, May 31, 2011 at 7:37am by Al

Here is what I came up with, but not sure if it is correct. 2 moles of NaOH = 79.994g NaOH 79.994g/1L * 1L/1000mL * 100mL/1g = 7.9
Thursday, January 10, 2013 at 6:08pm by James

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

Unless I have really missed something, and I don't think I have, there is no correct answer. There is no unknown to determine. 5.00 mL of 3 M HCl CAN'T require 14.45 mL of 1.00 M NaOH. moles = M x L. For HCl that is 0.005 x 3.00 M = 0.015 moles. NaOH titrated it. 0.01445 x 1....
Sunday, June 13, 2010 at 12:03am by DrBob222

General Chemistry
millimoles HCl = 8.00 mL x 0.1M=0.8mmols. mmoles NaOH = 2.50 mL x 0.1M = 0.25 mmols. mmoles NaOH = 9.50 x 0.1M = 0.95 mmoles. mmoles HAc = 8.00 mL x 0.1M = 0.8 mmoles. -------------------------------------- ..........HCl + NaOH ==> H2O + NaCl initial...0.8....0........0...
Sunday, November 6, 2011 at 9:28pm by DrBob222

College CHem
Suppose you accidently add too much NaOH during a titration. You then add a volume of HCl exactly equal to the volume of NaOH actually used in the titration. What effect does this have on the Ka?
Monday, March 15, 2010 at 6:43pm by Stephany Toussaint

I wonder WHICH titration end point was reached. 2NaOH + H2SO3 ==> Na2SO3 + 2H2O moles NaOH = M x L = ?? Use the coefficients in the balanced equation to convert moles NaOH to moles H2SO3. M = moles/L.
Thursday, October 7, 2010 at 9:18pm by DrBob222

1) H2Ma + NaOH --> H2O + NaHMa 2) NaHMa + NaOH --> H2O + Na2Ma I think is what they are after so you end up with the disodium salt.
Tuesday, March 3, 2009 at 2:58am by Dr Russ

If 33.0 mL of 0.002 M aqueous H3PO4 is required to neutralize 28.0 mL of an aqueous solution of NaOH, determine the molarity of the NaOH solution.
Sunday, April 22, 2012 at 3:22pm by Anonymous

HCl is an acid. NaOH is a base. general equation is as follows: an acid + a base = salt + water HCl + NaOH ==> NaCl + H2O
Sunday, September 29, 2013 at 12:39pm by DrBob222

ahhh im confused!! Calculate the pH of a solution formed by mixing 372 mL of a solution containing 4.4 x 10-6 M NaOH with 286 mL of 6.0 x 10-2 M NaOH. Report your answer to 2 decimal places.
Friday, November 16, 2007 at 6:14pm by mark

Convert 0.40 g NaOH to moles. Convert 50.0 mL x 0.1 M HCOOH to moles. See which is in excess. If NaOH, then calculate pOH and pH. If HCOOH, it will be a buffer and you should use the Henderson-Hasselbalch equation.
Wednesday, May 5, 2010 at 12:28pm by DrBob222

Calculate the theoretical pH after 2.50 mL and 9.50 mL of NaOH has been added in both the titration of HCl and of HC2H3O2. Indicate if the volume of NaOH is before or after the equivalence point.
Friday, May 9, 2008 at 9:45pm by paige

I saw your post last night and started to answer it before I went to bed; however, you don't have enough information to answer. You don't have the concn of the NaOH listed. So 10.00 mL of WHAT concn NaOH.
Thursday, October 27, 2011 at 10:15am by DrBob222

practice exam...not sure how to do this A solution contains 10.0 mmol of H3PO4 and 5.0 mmol of NaH2PO4. How many millimeters of 0.10 M NaOH must be added to reach the second equivilence point of the titration of the H3PO4 with NaOH?
Sunday, November 20, 2011 at 11:32am by Minx64

-A sample consist 90% NaOH and 10% CaO,whats the concentration of NaOH in the solution of this 3g sample,dissolved in 250cm3. -How many cm3 of H2SO4 with concentration of 0.255M are needed for titration of 100cm3 of this solution
Wednesday, September 4, 2013 at 8:28am by plsHelp

A 0.1mol/dm^3 aqueous solution of phosphoric (V) acid, H3PO4, is mixed with a 0.1 mol/dm^3 of aqueous solution of sodium hydroxide. Which mixture will form the salt Na3PO4? a) 10cm^3 of H3PO4 with 30cm^3 of NaOH b) 10cm^3 of H3PO4 with 10cm^3 of NaOH c) 20cm^3 of H3PO4 with ...
Friday, October 9, 2009 at 11:32pm by Marrion

No. CH3CH2CH2Br + NH3 ==> CH3CH2CH2NH3Br. I took a shortcut by using R to stand for CH3CH2CH2 and X to stand for Br to make RX + NH3 ==> RNH3Br. If you treat the CH3CH2CH2NH3Br with base (like NaOH), you will get the CH3CH2CH2NH2 + H2O + NaBr (in other words by treating ...
Tuesday, May 12, 2009 at 10:21pm by DrBob222

Write the equation. H2SO4 + 2NaOH ==> Na2SO4 + 2H2O step 2. Convert 50.00 mL and 0.135 M NaOH to mols NaOH. step 3. Convert mols NaOH to mols H2SO4 using the coefficients in the balanced equation. step 4. Convert mols H2SO4 to mL remembering that mols = L x M. You have mols...
Sunday, September 7, 2008 at 1:26pm by DrBob222

Weight of the mustard package Sample: 3.02 (g) Weight of the mustard package Solution: 33.3 (g) Trial #1 Trial #2 Trial #3 Weight of Mustard Package Solution Delivered (g) : 1.09 .948 .909 Weight of NaOH Solution Delivered (g) : .354 .304 .269 Concentration of NaOH : 7.48e-3...
Tuesday, October 16, 2012 at 10:32pm by Den

HC2H3O2 + NaOH ==> NaC2H3O2 + H2O How many moles is 1 g NaC2H3O2. That is g/molar mass = 1g/82 = 0.0122 moles. How can we get 0.0122 moles HC2H3O2. M = moles/L or L = moles/M = 0.0122/0.5 = 0.02440L or 24.40 mL of 0.5M HC2H3O2. Then dump in NaOH until all of it is ...
Wednesday, April 6, 2011 at 4:42pm by DrBob222

chem-acid-base titrations
ok for part 2 moles of NaOH=L*M .01*.5052=.005052 moles of NaOH then moles of NaOH =moles of HC2H3O2 since 1-1 ratio from equation that makes .005052 moles of HC2H3O2 correct? then M acetic acid=.005052 moles/.01688L which =2.9*10^-1 is that correct? and for part 3 is it mols ...
Monday, April 21, 2008 at 10:06pm by natash

Which of the following pairs of substances, when mixed in any proportion you wish, can be used to prepare a buffer solution? (Select all that apply.) NaCN and HCN NaCN and NaOH HCN and NaOH HCl and NaCN HCl and NaOH and Which of the following gives a buffer solution when equal...
Tuesday, April 12, 2011 at 7:04pm by Rebekah

moles acetic acid = M x L moles acetate = M x L. Add NaOH. mols added = M x L. NaOH reacts with acetic acid to produce more acetate at the expense of the acetic acid. Therefore, calulate moles NaOH added, subtract from acetic acid and add to acetate. Then M = moles/L, ...
Saturday, May 1, 2010 at 12:48pm by DrBob222

The original post says 25 mL of 2.00 M NaOH = 25 x 2 = 50 millimoles or 0.05 moles NaOH. But your solution says 0.2 x 0.025 = ??. Something is awry. Is it 2.00 M or 0.200 M? I think that's the trouble spot.
Monday, August 10, 2009 at 12:12pm by DrBob222

NaOBr, the salt of a strong base (NaOH) and a weak acid (HOBr), ionizes in water as follows: NaOBr(s) + H2O ==> NaOH + HBr and the net ionic equation is OBr^- + HOH ==> HOBr + OH^-
Sunday, April 18, 2010 at 9:43pm by DrBob222

Which of the following mixtures will result in the formation of a buffer solution when dissolved in 1.00 L of water? i) 0.50 mol NaOH and 0.50 mol HCl. ii) 0.50 mol NaCl and 0.25 mol HCl. iii) 0.50 mol NaF and 0.25 mol HF. iv) 0.50 mol NaOH and 0.25 mol HF. v) 0.25 mol NaOH ...
Friday, July 23, 2010 at 1:18pm by B

Chemistry New question
If 12.5 mL of 8.6 multiplied by 10-2 M NaOH is added to 32.5 mL of 6.0 multiplied by 10-4 M NaOH, what is the pH of the solution?
Monday, April 5, 2010 at 3:00pm by Jacob

How many mols HCl do you have? M x L = mols. Write the equation. How many mols NaOH do you need? mols NaOH = grams/molar mass. Solve for grams.
Friday, May 11, 2012 at 12:04am by DrBob222

You must recognize that when adding NaOH to HC2H3O2 (which is acetic acid and I will call it HAc and use Ac^- for acetate ion), you are forming a buffered solution. That means use the Henderson-Hasselbalch equation. HAc + NaOH ==> NaAc + H2O. moles HAc = 1.0 L x 2.0 M = 2 ...
Monday, April 12, 2010 at 9:18pm by DrBob222

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

Sulfuric acid reacts with sodium hydroxide according to this equation: H2S04 + 2 NaOH Na2(SO4) + 2 H2O A 10.00 mL sample of the H2SO4 solution required 13.71 mL of 0.309 M NaOH for neutralization. Calculate the molarity of the acid.
Wednesday, November 21, 2012 at 2:53pm by Liang9506

You need to know what indicator is being used. Methyl red indicator titrates the first H and phenolphthalein titrates the first and second H. The third one cannot be titrated easily. From the equations below, NaOH + H3PO4 ==> NaH2PO4 + H2O 2NaOH + H3PO4 ==> Na2HPO4 + ...
Saturday, July 9, 2011 at 9:50pm by DrBob222

A water containing 1x10^-4 mol CO2/L and having an alkalinity of 2.5x10^-4 eq/L has a pH of 6.7. The pH is to be raised to pH 8.3 with NaOH. How many moles of NaOH per liter of water are needed for this pH adjustment? (pK1=6.3 and pK2=10.3)
Tuesday, October 12, 2010 at 1:07pm by Ann

millimoles NaOH = 5 mL x 0.1M = 0.5 millimoles HAc = 25 mL x 0.2M = 5 ...........NaOH + HAc ==> NaAc + H2O initial....0.5.....5.0.....0.......0 change....-0.5....-0.5...+0.5...+0.5 equil.......0.....4.5......0.5....0.5 Substitute into the Henderson-Hasselbalch equation and ...
Thursday, September 8, 2011 at 11:57pm by DrBob222

Chemistry(Please help, thank you!)
ok so for mole of kHP it would be 1/204.44 = 0.00489 g/mol? The volume of NaOH for trial 1 was 14.50 mL and for trial 2 it was 9.90mL. So would it be 39.998 which is molecular mass of NaOH divided by the volume?
Monday, October 24, 2011 at 8:25pm by Hannah

So, Mols NaOH = 0.2264 M NaOH x 0.04936 L NaOH = 0.011175 Mols KHP = grams KHP / molar mass KHP or grams KHP = Mols KHP x molar mass KHP so 0.011175 x 204.22 = 2.2822 grams of KHP Is that correct DBob222?
Saturday, December 1, 2012 at 9:16am by T

It all depends upon when the air bubbles were there, if they changed in size during the titration, or if they were dispensed in the titration. Go back to the NaOH/H2C2O4 titration did, work up the formulas used to calculate the M of the NaOH and the M H2SO4 and go through the ...
Saturday, April 2, 2011 at 11:58am by DrBob222

H2SO4 + 2NaOH ==> Na2SO4 + 2H2O 0.59 mol NaOH x (1 mol H2SO4/2 mol NaOH) = ?
Thursday, January 19, 2012 at 2:40am by DrBob222

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation: Al2O3(s) + NaOH(l) + HF(g) -> Na3AlF6 + H2O(g) If 14.0 kilograms of Al2O3(s), 57.4 kilograms of NaOh(l), and 57.4 kilograms of HF(g) react ...
Sunday, September 18, 2011 at 9:48am by Sandra

An unknown solid acid is either citric acid or tartaric acid. To determine which acid you have, you titrate a sample of the solid with NaOH. The appropriate reactions are as follows: Citric acid: H3C6H5O7 (aq) + 3NaOH (aq) -> 3H2O (l) + Na3C6H5O7 (aq) Tartaric acid: ...
Tuesday, May 7, 2013 at 1:25pm by Sarah

pH = 10 pOH = 14-10 = 4 pOH = -log(OH^-) (OH^-) = 1 x 10^-4 Since NaOH is a strong base, then we want a concn of NaOH of 1 x 10^-4 M.
Wednesday, April 16, 2008 at 9:26am by DrBob222

Chemistry using concentration
mols NaOH = grams/molar mass Calculate mols NaOH needed. Then M = mols/L You have mols and M, calculate L.
Monday, May 12, 2008 at 12:37am by DrBob222

lets say we have Cu(NO3)2 and then we add NaOH. so Cu(NO3)2 + NaOH and then we add HNO3 to it what wil happen?
Sunday, March 21, 2010 at 11:38pm by Dalala

Chemistry need response ASAP please
mols NaOH = M x L = ? mols KHC8H4O4 = mols NaOH. grams KHC8H4O4 = mols x molar mass.
Wednesday, March 28, 2012 at 10:52pm by DrBob222

10L of .15M solution contains 1.5 moles NaCl. Each mole of NaCl produces 1 mole of NaOH So, how many grams in 1.5 moles NaOH?
Wednesday, April 17, 2013 at 1:36am by Steve

If 0.250 moles of NaOH are dissolved in 125g of water, what is the %m/m of NaOH in this solution? I converted 0.250 mol to 10.0 g. Then using the equation mass solute/mass solution * 100 I came up with the final answer of 7.41 %m/m. Is that correct?
Tuesday, April 29, 2008 at 2:15pm by Jen!! =)
MaVa = MbVb can be used for the reaction between HCl and NaOH. Can it be used for the reaction between NaOH and H2C2O4 (oxalic acid, which can be found in rhubarb leaves, contains 2 acidic protons)? Explain your answer showing the balanced equation
Wednesday, November 17, 2010 at 9:24pm by Stacy

chem lab
Calculate the theoretical pH after 2.50 mL and 9.50 mL of NaOH has been added in both the titration of HCl and of HC2H3O2. Indicate if the volume of NaOH is before or after the equivalence point. i dont know where to start.
Tuesday, May 10, 2011 at 2:00am by Anonymous

Chemistry(Please help, thank you!!)
Determine the concentration of a NaOH solution if 27.80mL of NaOH is required to neutralize 10.00mL of a 1.00 M H2SO4 solution. (Hint:Review monoprotic vs. diprotic acids.) would I do (27.80)(10)? Thank you for your help!!
Tuesday, October 25, 2011 at 10:21pm by Hannah

In a titration, 3.4 g of an acid (HX) requires 21.5 mL of 0.85 M NaOH(aq) for complete reaction. What is the molar mass of the acid? Answer in units of g/mol. im not really sure how to even start... but i calculated the moles of NaOH which i got to be .018 moles.... what do i ...
Sunday, October 24, 2010 at 5:40pm by Lulu

Sodium oxide, reacts with water to give NaOH. A) Write a balance equation? B)What is the pH of the solution prepared by allowing 1.55 g of Na2O to react with 500.0 ml of water? (assume there is no volume change) C) How many milliliters of 0.0100M HCL are needed to neutralize ...
Sunday, November 21, 2010 at 5:03pm by Ann

NaOH + HBr ==> NaBr + H2O moles NaOH used in the titration = M x L = 0.37 x 0.0878 = 0.03249. Using the balanced equation above, convert moles NaOH to moles HBr. The coefficients are 1:1; therefore, moles HBr = 0.03249 That = the moles in the 102.1 mL sample. How much was ...
Sunday, May 1, 2011 at 11:00pm by DrBob222

That's 20.5g NaOH in 100 g soln. The 100 g soln is composed of 20.5g NaOH + 100-20.5 ==> 79.5 g H2O Convert 20.5g NaOH to moles. Convert 79.5 g H2O to moles. XNaOH = moles NaOH/total mols. XH2O = moles H2O/total mols.
Sunday, February 26, 2012 at 7:29pm by DrBob222

OK. Here is what you do. Chemistry works by mols. Mols of X react with mols Y to produce mols Z + mols w. The definitioin of molarity is # mols/Liter of solution. So we go with that definition, solving for # mols and that give #mols = M x L. So how many mols HCl to we have? We...
Wednesday, May 7, 2008 at 9:40pm by DrBob222

1. When performing this experiment, a student mistakenly used impure KHP to standardize the NaOH solution. If the impurity is neither acidic nor basic, will the percent by mass of acetic acid in the vinegar solution determined by the student be too high or too low? Justify ...
Sunday, March 23, 2014 at 3:05pm by anonymous

A sample of solid monoprotic acid with molar mass equal to 169.7 g/mol was titrated with 0.1599 M sodium hydroxide solution. Calculate the mass in grams of acid to be used if the volume of NaOH to be used is 25 mL. Is the solution correct?: moles NaOH = 0.1599 x 0.025=3.998x10...
Friday, October 3, 2008 at 12:17am by Erika

A clarifying note: The NaOH is a solid and HCl is a gas. They will not react in that state. Most of the time we omit the fact that we use solutions but if we want to do it up brown, we would write HCl(aq) + NaOH(aq) ==> NaCl(aq) + H2O(l) Thanks for using Jiskha.
Friday, December 19, 2008 at 10:10am by DrBob222

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

acid-base titrations
For each of the following circumstances, indicate whether the calculated molarity of NaOH would be lower, higher or unaffected. Explain your answer in each case. a.the inside of the pipet used to transfer the standard HCI solution was wet with water. b.You added 40mL of water ...
Monday, April 21, 2008 at 10:22pm by natash

Sodium metal reacts violently with water to form NaOH and release hydrogen gas. Suppose that 3.00 g of Na reacts completely with 3.00 L of water and the final volume of the system is 3 L. What is the molarity M of the NaOH Solution formed by the reaction?
Wednesday, May 1, 2013 at 8:44am by Janet

organic chemistry
An unknown H2So4 sample of 25.00mL vl is titrated with 0.149 M NaOH. If 45.22mL of NaOH were needed to reach the pink endpoint, what was the molarity of the unknown acid? Show work using conversion factors????? H2SO+2NaOH --> Na2SO4+2H2O Dont know where or how to start!
Tuesday, January 31, 2012 at 10:01pm by April

16.7 m means 16.7 mol NaOH/kg solvent. 16.7 mol NaOH = mols x molar mass = about 670 g but you need to this more accurately. Thus 16.7 m contains about 670g NaOH/kg solvent. That is 670g/1000 g solvent. The solution, then, must weigh 1000g + 670 g = 1670g. mass% = (670/1670)*...
Sunday, January 13, 2013 at 5:49pm by DrBob222

Question 26A student completes a titration by adding 12.0 milliliters of NaOH(aq) of unknown concentration to 16.0 milliliters of 0.15 M HCl(aq). What is the molar concentration of the NaOH(aq)?
Monday, May 3, 2010 at 8:35pm by frnk

the density of a certain aqueous solution is 1.17g/ml, and the solution is 3.57% by mass NaOH. how many mL of this solution would you need to use in order to prepare 100.0 mL of .150 M NaOH solution?
Monday, September 6, 2010 at 6:46pm by zahra

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

Pages: <<Prev | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10