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Search: 0.050 M propanoic acid (HC3H5O2 Ka = 1.30 × 10−5) / 0.080 M sodium propanoate after the addtion of 0.069 mol solid NaOH

Number of results: 116,489

chemistry
Calculate the pH of each of the following solutions. (a) 0.300 M propanoic acid (HC3H5O2, Ka = 1.3 10-5) (b) 0.300 M sodium propanoate (NaC3H5O2) (c) pure H2O (d) 0.300 M HC3H5O2 and 0.300 M NaC3H5O2
Friday, December 2, 2011 at 2:44pm by melissa

Chemistry
Calculate the pH after 0.011 mol NaOH is added to 1.00 L of each of the four solutions. a) 0.100 M propanoic acid (HC3H5O2, Ka= 1.3 10-5) b) 0.100 M sodium propanoate (NaC3H5O2) c)pure H2O d) 0.100 M HC3H5O2 and 0.100 M NaC3H5O2
Thursday, March 15, 2012 at 10:03pm by Mike

Chemistry
What is the pH of a 1.00 L solution of the following buffer 0.500 M propanoic acid (HC3H5O2 Ka = 1.30 × 10^−5) / 0.800 M sodium propanoate after the addition of 0.700 mol solid NaOH. I tried using the Henderson-Hasselbach equation pH = pKa + log[A-/HA] I used it in a ...
Tuesday, October 27, 2009 at 7:32pm by Anonymous

Chemistry
What is the pH of a solution containing 0.10 M propanoic acid (HC3H5O2) and 0.10 M sodium propanoate after 0.02 mol of HCl is added to 1.0 L of the solution?
Sunday, February 28, 2010 at 8:39pm by Blake

chemistry
6.25 L of a propanoic acid-propanoate buffer must be prepared. The pH of the buffer needs to be 5.05. The solution will be 0.120M in propanoic acid. If the propanoate ion comes from the strontium propionate Sr(C3H5O2)2 then how many grams of strontium propanoate are needed?
Friday, March 1, 2013 at 10:37am by Studious

Chemistry
Need help with AP chemistry, specifically Acids and Bases The acid ionization constant, Ka, for propanoic acid, C2H5COOH, is 1.3  10 -5 . (a) Write the expression for the acid-dissociation constant, Ka for propanoic acid. (b) Calculate the hydrogen ion concentration, [...
Monday, April 9, 2012 at 8:59pm by Sam

ap chemistry
For propanoic acid (HC3H5O2, Ka = 1.3 10-5), determine the concentration of all species present, the pH, and the percent dissociation of a 0.29 M solution. [H+] [C3H5O2-] [OH -] [HC3H5O2] pH percent dissociation
Wednesday, April 6, 2011 at 8:16pm by Ryu

chemistry
propanoic acid ch3ch2cooh is a weak acid which has a value for dissociation content of ka = 6.3 x 10^-6 calculate the pH of an aqueous solution of propanoic acid containing 37 g of propanoic acid per litre. i'm stumbled on this one.. i calculated that the concentration of the ...
Tuesday, July 27, 2010 at 9:45pm by rick

CHEMISTRY
For propanoic acid (HC3H5O2, Ka = 1.3 10-5), determine the concentration of all species present, the pH, and the percent dissociation of a 0.170 M solution. [HC3H5O2]: [C3H5O2‾]: [H+] : [OH ‾]: pH : i need help on how to find the concentration of these also in Moles.
Tuesday, November 23, 2010 at 4:38pm by AMANDIP

Chem
For a solution that is 0.280 M HC3H5O2 (propanoic acid Ka=1.3*10^-5) and 0.0894 M HI, calculate the following. Concentration of H30, OH, C3H5O2-, I-
Thursday, April 14, 2011 at 4:42pm by Joe

Chemistry, #5
For propanioic acid, HC3H5O2, Ka=1.3 x 10^-5, determine the concentration of the species present, the pH and the percent dissociation of a 0.21 M solution. Do this for H+, OH-, C3H5O2-, HC3H5O2, and the pH and percent dissociation. If someone can please just help me get ...
Saturday, August 4, 2007 at 4:50pm by Taasha

CHEMISTRY
For propanoic acid (HC3H5O2, Ka = 1.3 10-5), determine the concentration of all species present, the pH, and the percent dissociation of a 0.170 M solution. PLEASE HELP ME DO THIS PROBLEM.
Tuesday, November 23, 2010 at 4:38pm by AMANDIP

Chemistry
You make 1.00 L of a buffered solution (pH = 5.10) by mixing propanoic acid and potassium propanoate. You have 1.00 M solutions of each component of the buffered solution. What volume of each solution do you mix to make such a buffered solution? propanoic acid potassium ...
Wednesday, March 13, 2013 at 2:53am by Margaret

Chemistry
Use the henderson-hasselbalch equation pH=pKa+log[A-/HA] where Pka=-logKa Ka=1.3 x 10–5 HA=propanoic acid A-=conjugate base of propanic acid 0.0300L *(0.165M)=moles of propanoic acid 0.010*(0.300M)= moles of KOH 0.005*(0.300M)= moles of KOH Solve for moles of conjugate base of...
Tuesday, March 19, 2013 at 8:22pm by Devron

chemistry
Consider the Ka values for the following acids: Cyanic acid, HOCN, 3.5 ´ 10-4 Formic acid, HCHO2, 1.7 ´ 10-4 Lactic acid, HC3H5O3, 1.3 ´ 10-4 Propionic acid, HC3H5O2, 1.3 ´ 10-5 Benzoic acid, HC7H5O2, 6.3 ´ 10-5 Which of the following is the weakest base? is it propanionic i ...
Wednesday, May 26, 2010 at 1:31am by michael

Chemistry! Please Help!
Let's call propionic acid HP, then it will ionize as HP ==> H^+ + P^- First convert pH = 2.50 to (H^+) by pH = -log(H^+). Then set up the expression for Ka. Ka = (H^+)(P^-)/(HP). You know (H^+) now from the pH calculation. You know (P^-) because it equals the same as (H...
Thursday, February 12, 2009 at 7:46pm by DrBob222

chemistry
The bromination of acetone is acid-catalyzed. CH3COCH3 + Br2 CH3COCH2Br + H+ + Br - The rate of disappearance of bromine was measured for several different concentrations of acetone, bromine, and H+ ions at a certain temperature. [CH3COCH3] [Br2] [H+] rate of disappearance of ...
Monday, February 1, 2010 at 1:22pm by danielle

Chemistry
Can someone please help me answer these questions? Is additional information other than what is given needed to solve? Thanks in advance. 1) Calculate the PH of a buffer containing 0.100 M propanoic acid, HC3H5O2 and .100 M NaC3H502 after the following have been added. Ka for ...
Friday, December 13, 2013 at 1:39am by Charles

chemistry
For 0.100 M of weak acid (propanoic acid HPr) the Ka is given by 1.30 x10-5 , calculate the pH of the sample.
Saturday, April 18, 2009 at 5:11pm by lil'mama

Chemistry
What must be the molarity of an acetic acid solution if it has the same percent ionization as 0.150M HC3H5O2 propionic acid,Ka=1.3*10^-5 )?
Thursday, February 16, 2012 at 11:34pm by Tiana

General Chemistry
Calculate the pH and include the balanced equation for the acid dissociation reaction. 0.20 M HC3H5O2 (propionic acid, Ka = 1.3 x 10^-5)
Monday, July 29, 2013 at 3:55am by <3

chemistry
pls help me. A buffer was prepared by dissolving 0.100 mol of the weak acid HA (Ka = 1.00 × 10−5) plus 0.050 mol of its conjugate base Na+A− in 1.00 L. Find the pH.
Saturday, December 1, 2012 at 4:09pm by canas

AP Chem
I'm lazy so let's call sodium propanoate NaPr. The Pr anion is hydrolyzed. Pr^- + HOH ==> HPr + OH^- Kb= (Kw/Ka) = (HPr)(OH^-)/(Pr^-) Substitute for Kw and Ka (for propionic acid) and (Pr^-). Let X = HPr and OH. Solve for x, convert to pOH, then to pH.
Wednesday, March 9, 2011 at 10:16pm by DrBob222

chemistry
Sodium propanoate we will call NaP. It's the propanoate ion that is hydrolyzed in solution. .......P^- + HOH ==> HP + OH^- I.....0.3............0....0 C......-x............x....x E.....0.3-x..........x....x Kb for P^- = (Kw/Ka for HP) = (x)(x)/(0.3-x) Solve for x = (OH^-) ...
Saturday, July 20, 2013 at 1:36pm by DrBob222

Chemistry
Propanoic acid has a Ka of 1.3 x 10^-5. What is the % ionization in a 3.0M solution?
Tuesday, May 22, 2012 at 8:40am by Lily

chemistry
For a solution that is 0.280M HC3H5O2 (propionic acid, Ka= 1.3*10^-5) and 0.0894M HI , calculate the following: [H3O+] [OH-] [C3H5O2-] [I-]
Thursday, April 14, 2011 at 7:38am by Saunders

chemistry
One buffer solution used to calibrate a pH electrode was made by dissolving 2.24g of potassium propanoate, C2H5COOK, in 250cm3 of 0.20 mooll-1 propanoic acid.
Wednesday, October 24, 2012 at 8:20pm by Jen

chemistry
Which weak acid would be best to use when preparing a buffer solution with a pH of 8.50? 1) an acid with Ka=3.3x10^-9 2) an acid with Ka=5.0x10^-4 3) an acid with KA=6.3x10^-6 4) an acid with Ka=5.0x10^-7 5) an acid with Ka=4.0x10^-5 6) an acid with Ka=2.0x10^-10
Monday, April 8, 2013 at 3:31pm by Anonymous

chem--need help, please!!!
Ka for benzoic acid, C6H5COOH, 6.5x10^-5. Calculate the pH of solution after addition of 10.0, 20.0, 30.0, and 40.0 mL of 0.10 M NaOH to 40.0 mL of 0.10 M Benzoic acid. PLEASE CHECK MY ANSWER!!!!! My answer is: Moles acid = 0.040 L x 0.10 M = 0.0040 Moles base = 0.010 L x 0.10...
Wednesday, October 20, 2010 at 12:33pm by Andy

chemistry
pls help i need the answer not jus part of it, pls im really stuck out here. A buffer was prepared by dissolving 0.100 mol of the weak acid HA (Ka = 1.00 × 10−5) plus 0.050 mol of its conjugate base Na+A− in 1.00 L. Find the pH.
Saturday, December 1, 2012 at 4:45pm by canas

Chemistry
A 50.0 mL sample of 0.23 M propanoic acid, CH3CH2COOH, a weak monoprotic acid, is titrated with 0.14 M KOH. Ka of CH3CH2COOH = 1.4 multiplied by 10-5. What is the pH at equivalence point?
Sunday, April 3, 2011 at 5:10pm by George

chemistry
An unknown weak acid with a concentration of 0.32 M was found to have a pH of 2.7. Using the table of dissociation constants (Ka) below, determine what the likely identity of the acid is. Explain how you got your answer (hint - calculate the Ka in the same way you did for ...
Monday, March 25, 2013 at 12:48am by Anonymous

CHEMISTRY HELP NEEDED!!
Given info: Reaction: 3A+2B--->2C+D {A] (mol/L) 1.0 ×10^-2 1.0 ×10^−2 2.0 ×10^−2 2.0 ×10^−2 3.0 ×10^−2 [B] (mol/L) 1.0 3.0 3.0 1.0 3.0 Rate of appearance of C (mol/L-hr) 0.30×10^−6 8.10×10^−6 3.24×10^−5 1.20×10^−6 7.30×10^&#...
Sunday, March 11, 2012 at 9:00pm by Kelly

Chemistry
Given info: A] (mol/L) [B] (mol/L) Rate of appearance of C (mol/L-hr) 1.0 ×10^-2 1.0 0.30×10^−6 1.0 ×10^−2 3.0 8.10×10^−6 2.0 ×10^−2 3.0 3.24×10^−5 2.0 ×10^−2 1.0 1.20×10^−6 3.0 ×10^−2 3.0 7.30×10^−5 Reaction: 3A+2B--->...
Sunday, March 11, 2012 at 8:47pm by Kelly

chemistry
A lab technician tests a 0.100 mol/L solution of propanoic acid and finds that its hydrogen ion concentration is 1.16 x 10^-3 mol/L. Calculate the percent ionization of propanoic acid in water
Saturday, April 5, 2008 at 6:36pm by Geroge B.

Chemistry
You will need the Ka for sulfurous acid to calculate the pH as it is a weak acid. Were you given a figure in the question? H2SO3<> HSO3- + H+ Ka = 1.54×10−2 mol L^-1 (from a data book) at start H2SO3 HSO3- H+ 0.250M 0M 0M at equilibrium 0.250-x x x Ka=[HSO3-][H+]/[...
Friday, February 18, 2011 at 2:51am by Dr Russ

General Chemistry 1412 inorganic
Find the pH of a solution that is 0.30 M benzoic acid (HBz) and 0.25 M sodium benzoate(NaBz). Ka for benzoic acid = 6.5 x 10-5
Wednesday, April 18, 2012 at 9:33pm by generalchem1412

chemistry
calculate the Ph of 0.3 M sodium propanoate, Ka 2.0 x 10 base -3. then calculate the pH after 0.03 mol NCL is added to 2L of the first solution.
Saturday, July 20, 2013 at 1:36pm by Mandy

College Chemistry
"What is the pH of a solution that is 0.50 M in sodium acetate and 0.75 M in acetic acid? (Ka for acetic acid is 1.85 x 10-5)." Would i use the Ka expression formula??
Tuesday, October 19, 2010 at 11:02pm by James

chem
What are the equilibrium concentrations of H2SO3, H+, HSO3−, and SO32−in a 0.050 M solution of sulfurous acid H2SO3 at 25 oC? For H2SO3 at 25 oC, Ka1 = 1.5×10−2 and Ka2 = 1.0×10−7
Monday, February 25, 2013 at 6:07am by ston

chem
What are the equilibrium concentrations of H2SO3, H+, HSO3−, and SO32−in a 0.050 M solution of sulfurous acid H2SO3 at 25 oC? For H2SO3 at 25 oC, Ka1 = 1.5×10−2 and Ka2 = 1.0×10−7
Monday, February 25, 2013 at 7:10am by ston

chemistry
The pH of a solution that contains 0.818 M acetic acid (Ka = 1.76 x 10-5) and 0.172 M sodium acetate is __________. The Ka of acetic acid is 1.76 × 10-5.
Tuesday, November 23, 2010 at 11:22pm by jack

chemistry
The pH of a solution that contains 0.818 M acetic acid (Ka = 1.76 x 10-5) and 0.172 M sodium acetate is __________. The Ka of acetic acid is 1.76 × 10-5.
Wednesday, November 24, 2010 at 10:22am by jack

chemistry
The pH of a solution that contains 0.818 M acetic acid (Ka = 1.76 x 10-5) and 0.172 M sodium acetate is __________. The Ka of acetic acid is 1.76 × 10-5.
Wednesday, November 24, 2010 at 12:44pm by jack

Chemistry
Given that the Ka of benzoic acid is 6.50 x 10-5, how would one prepare 0.500 L of a benzoic acid/sodium benzoate buffer of a desired pH of 4.00? The starting concentration of the benzoic acid is 3.00 M and the molecular weight of sodium benzoate is 144.10 g/mol.
Sunday, April 10, 2011 at 7:48pm by Nadine

Chemistry
Given that the Ka of benzoic acid is 6.50 x 10-5, how would one prepare 0.500 L of a benzoic acid/sodium benzoate buffer of a desired pH of 4.00? The starting concentration of the benzoic acid is 3.00 M and the molecular weight of sodium benzoate is 144.10 g/mol
Sunday, April 10, 2011 at 7:48pm by Mike

Chemistry
Given that the Ka of benzoic acid is 6.50 x 10-5, how would one prepare 0.500 L of a benzoic acid/sodium benzoate buffer of a desired pH of 4.00? The starting concentration of the benzoic acid is 3.00 M and the molecular weight of sodium benzoate is 144.10 g/mol
Sunday, April 10, 2011 at 9:36pm by Mike

Diana’s dresses: the story behind the fashio
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Sunday, April 28, 2013 at 6:33pm by Karlie Kloss: I&amp;amp;amp;amp;amp;amp;amp;am

Chemistry
Calculate the pH at the following points during the titration of 100.0ml of 0.20 M acetic acid (ka for acetic acid=1.8*10^-5) with 0.10 M sodium hydroxide. 1. Before addition of any base 2. after addition of 30.0 mL of base
Wednesday, April 20, 2011 at 5:29pm by Phill

Chemistry
calculate the specific rate constant A] (mol/L) [B] (mol/L) Rate of appearance of C (mol/L-hr) 1.0 ×10^-2 1.0 0.30×10^−6 1.0 ×10^−2 3.0 8.10×10^−6 2.0 ×10^−2 3.0 3.24×10^−5 2.0 ×10^−2 1.0 1.20×10^−6 3.0 ×10^−2 3.0 7.30×10^−5
Sunday, March 11, 2012 at 5:41pm by Kelly

Chemistry
calculate the specific rate constant A] (mol/L) [B] (mol/L) Rate of appearance of C (mol/L-hr) 1.0 ×10^-2 1.0 0.30×10^−6 1.0 ×10^−2 3.0 8.10×10^−6 2.0 ×10^−2 3.0 3.24×10^−5 2.0 ×10^−2 1.0 1.20×10^−6 3.0 ×10^−2 3.0 7.30×10^−5
Sunday, March 11, 2012 at 6:12pm by Kelly

Chemistry
calculate the specific rate constant A] (mol/L) [B] (mol/L) Rate of appearance of C (mol/L-hr) 1.0 ×10^-2 1.0 0.30×10^−6 1.0 ×10^−2 3.0 8.10×10^−6 2.0 ×10^−2 3.0 3.24×10^−5 2.0 ×10^−2 1.0 1.20×10^−6 3.0 ×10^−2 3.0 7.30×10^−5
Sunday, March 11, 2012 at 6:13pm by Kelly

Chemistry
calculate the specific rate constant A] (mol/L) [B] (mol/L) Rate of appearance of C (mol/L-hr) 1.0 ×10^-2 1.0 0.30×10^−6 1.0 ×10^−2 3.0 8.10×10^−6 2.0 ×10^−2 3.0 3.24×10^−5 2.0 ×10^−2 1.0 1.20×10^−6 3.0 ×10^−2 3.0 7.30×10^−5
Sunday, March 11, 2012 at 6:14pm by Kelly

Chemistry
You don't get pH from Ka UNLESS you have the concentration of the acid. In this case you don't have that so you can't get a straight pH. If you are referring to the first question, which is The following questions refer to the following acids and bases. A) Hydrazoic acid HN3 ...
Sunday, April 6, 2014 at 12:10pm by DrBob222

chemistry
What would be the pH of a solution of hy- poiodous acid (HOI) prepared by dissolving 144 grams of the acid in 200 mL of pure water (H2O)? The Ka of hypoiodous acid is 2¡¿10−11 1. 7 2. 10 3. 13 4. 5 5. 1
Thursday, February 25, 2010 at 10:44pm by hello

chem
make 300.0 mL of a pH 5.30 buffer from 0.500 M acetic acid and 0.500 M sodium acetate. Calculate the volume of 0.500 M acetic acid and the volume of 0.500 M sodium acetate required to make 300.0 mL of a buffer solution with pH = 5.30. (The Ka for acetic acid at 298 K is 2.00 ...
Tuesday, April 26, 2011 at 1:57am by Journey

chem
make 300.0 mL of a pH 5.30 buffer from 0.500 M acetic acid and 0.500 M sodium acetate. Calculate the volume of 0.500 M acetic acid and the volume of 0.500 M sodium acetate required to make 300.0 mL of a buffer solution with pH = 5.30. (The Ka for acetic acid at 298 K is 2.00 x...
Tuesday, April 26, 2011 at 2:16am by mie

Chemistry
Classify the following 3 acids in order of increasing acidity: Acid 1:Ka = 8 x 10-6
 Acid 2:Ka = 6 x 10-4
 Acid 3:Ka = 9 x 10-11
Sunday, May 13, 2012 at 9:32pm by Andrea

Chemistry
Calculate the pH of the solution formed by adding 15.0ml of 0.1M sodium hydroxide to 30.0ml of 0.1M propanoic acid.
Sunday, January 19, 2014 at 6:33pm by Fritz

Chemistry
This is worked the same way as the propanoic acid problem below; however, I really don't believe the Ka value. I thought HBrO3 was a strong acid (ionized 100%). The literature says that is so.
Tuesday, May 22, 2012 at 8:41am by DrBob222

chemistry
When 50.0 mL of 0.050 M formic acid, HCHO2, is titrated with 0.050 M sodium hydroxide, what is the pH at the equivalence point?
Wednesday, April 11, 2012 at 6:02am by chamy

Chemistry 122
A solution is prepared to be 0.10 M acetic acid, HC2H3O2, and 0.20 M sodium acetate, NaC2H3O2. What is the pH of this buffer? Ka for acetic acid is 1.7 x 10^-5. What is the pH after 9.5 mL of 0.10 M hydrochloric acid is added?
Monday, November 22, 2010 at 10:22am by Marlynda

chemsitry
buffer calculate the ph of the .1m propanoic acid Ka 1.35e-5
Sunday, April 10, 2011 at 9:09pm by neee

Chemistry
First, the question I'm given is: What is the pH of a 1.0 L solution containing 0.25M acetic acid and 0.75M sodium acetate ( Ka for acetic acid= 1.8x 10-5) So I took the -log(1.8x10-5)= 4.74+log(.75/.25)= 5.22 pH But then I'm asked if .050 mol NaOH is added to the above ...
Monday, April 8, 2013 at 7:01pm by Marcus

Urgent Chem Help
Question.. Nicotinic acid (niacin) is a monoprotic acid with the formula HC6H4NO2. A solution that is 0.012M in the nicotinic acid has a pH of 3.39 at 25C. what is the acid-ionization constant, Ka and pKa for this acid at 25C? Attempt.. HC6H4NO2 <----> C6H4NO2- + H+ [H...
Sunday, March 15, 2009 at 3:34pm by Saira

Plz -- Chem Help
Question.. Nicotinic acid (niacin) is a monoprotic acid with the formula HC6H4NO2. A solution that is 0.012M in the nicotinic acid has a pH of 3.39 at 25C. what is the acid-ionization constant, Ka and pKa for this acid at 25C? Attempt.. HC6H4NO2 <----> C6H4NO2- + H+ [H...
Sunday, March 15, 2009 at 3:35pm by Rushi

Chemistry
The following questions refer to the following acids and bases. A) Hydrazoic acid HN3 Ka = 1.9 x 10^-5 B) Hydrofluoric acid HF Ka = 6.8 x 10 ^-4 C) Nitrous Acid HNO2 Ka = 4.5 x 10^-4 D) Phenol C6H5OH Ka = 1.3 x 10^-10 E) Aniline C6H5NH2 Ka = 4.3 x 10^-10 1) which compound is ...
Sunday, April 6, 2014 at 12:10pm by Zachary

AP CHEMISTRY
Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 0.080 M NaOH. A) benzoic acid (C6H5COOH) Ka= 6.3*10^-5 for it
Thursday, March 1, 2012 at 4:26pm by DAN

chemistry
oh This is an acid base experiment this is the last stage. There are two parts: part 1) 0.3470g sodium acetate trihydate is added to 15.00 ml acetic acid (concentration is 0.20) then its mixed and a ph is taking--i got 4.30 PH part II) Added 0.7618 g of sodium acetate ...
Thursday, July 16, 2009 at 7:35pm by Jim_R

chemistry
Let's call propanoic acid HP. HP ==> H^+ + P^- Ka = (H^+)(P^-)/(HP) If 0.1 M HP acutaly has 1.16 x 10^-3 (H^+), then ionization = 1.16 x 10^-3/0.1 and you can convert that to percent by multiplying by 100.
Saturday, April 5, 2008 at 6:36pm by DrBob222

Chemistry
Let's call propanoic acid HPr. Then ...........HPr ==> H^+ + Pr^- Initial....3.0......0.....0 Change......-x......x......x Equil.....3.0-x.....x.......x Ka = (H^+)(Pr^-)/(HPr) Substitute from the ICE chart into the Ka expression and solve for x = (H^+). Then %ionization...
Tuesday, May 22, 2012 at 8:40am by DrBob222

Chemistry
1. Calculate the pH of a buffer solution that contains 0.32 M benzoic acid (C6H5CO2H) and 0.17 M sodium benzoate (C6H5COONa). [Ka = 6.5 × 10-5 for benzoic acid] Round your answer to two places past the decimal. 2. A solution is prepared by mixing 470 mL of 0.18 M Tris·Base and...
Monday, April 8, 2013 at 1:05am by Erin

Chemistry
1. Calculate the pH of a buffer solution that contains 0.32 M benzoic acid (C6H5CO2H) and 0.17 M sodium benzoate (C6H5COONa). [Ka = 6.5 × 10-5 for benzoic acid] Round your answer to two places past the decimal. 2. A solution is prepared by mixing 470 mL of 0.18 M Tris·Base and...
Monday, April 8, 2013 at 1:31am by Erin

chemistry
Benzoic Acid (HC6H5OO, 122.0 g/mol) has a Ka = 6.3 x 10-5. What is the pH of a solution that has 3.050 g of benzoic acid in enough water to make a 0.250 L solution?
Saturday, March 31, 2012 at 9:59pm by Anonymous

Chemistry
What is the PH of a buffer made by adding 8.20 grams of sodium acetate(NaCH3CO2) to 500 ml of 0.10 M acetic acid(CH3CO2H). Ka for acetic acid is 1.8 x 10^-5.
Sunday, April 14, 2013 at 8:17pm by Lele

Chem
What is the pH of the resulting solution if 30.00 mL of 0.100M acetic acid is added to 10.00mL of 0.100 M NaOH? For acetic acid, Ka=0.000018. I know that NaOH is a strong base and acetic acid is a weak acid. What is the equation: C6H5COOH+NaOH<--> C6H5COOH Once I get ...
Wednesday, February 27, 2008 at 4:22pm by Sarah

chemistry(Acid&Base)
Calculate the solubility product of calcium hydroxide if the solubility of Ca(OH)2(s) in water at 25◦C is 0.011 M. Choices: 1. 1.1 × 10−5 2. 5.3 × 10−6 3. 2.7 × 10−6 4. 1.5 × 10−8 5. 1.2 × 10−4
Thursday, May 12, 2011 at 6:05pm by vivian

BBC News
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Saturday, April 28, 2012 at 7:17pm by Karlie Kloss: I&amp;amp;amp;amp;amp;amp;amp;am

Chemistry
A buffer is prepared by mixing 0.30 mole of formic acid (HCNO2) and 0.20 mole of sodium formate (NaCHO2) in enough water to make 0.50 liter of solution. Ka of formic acid is 1.7 x 10-4. Calculate the pH of the resulting buffer solution.
Wednesday, March 7, 2012 at 9:18pm by Sam

Biochem
calculate the volumes of 0.10 M acetic acid and 0.20 M sodium acetate that are needed to prepare 48-mL buffer solution with pH = 3.70. Ka for acetic acid is 1.75 x 10-5
Friday, April 22, 2011 at 6:01am by Iris

Chemistry- Please Help!!
Acetic acid has a Ka = 1.75 x 10-5 and formic acid has a Ka = 1.8 x 10-4. Which is the stronger acid?
Wednesday, July 31, 2013 at 1:17pm by Kelly

Chemistry
* Fixed Question What is the PH of a buffer made by adding 8.20 grams of sodium acetate(NaCH3CO2) to 500 ml of 0.10 M(molarity) acetic acid(CH3CO2H). Ka for acetic acid is 1.8 x 10^-5.
Sunday, April 14, 2013 at 8:24pm by Lele

chemistry
The Ka of a monoprotic weak acid is 7.25 × 10-3. What is the percent ionization of a 0.167 M solution of this acid? If the concentration is more than 1000 times greater than the Ka, then you can make an assumption that simplifies the algebra needed to solve the problem. ...
Wednesday, April 10, 2013 at 11:57am by sara

CHEMISTRY
I don't like to type all that stuff, so propanoic acid I will call HP. HP ==> H^+ + P^- Ka = (H^+)(P^-)/(HP) H^+ = x P^- = x HP = 0.170-x Substitute into Ka and solve for x, then convert to pH. % diss = [(H^+)/0.170]*100 = ?? Post your work if you get stuck.
Tuesday, November 23, 2010 at 4:38pm by DrBob222

Chemistry
Benzoic acid (C6H5COOH)is a monoprotic weak acid with Ka=6.30*10^-5. What is the pH of a solution of benzoic acid that is 0.559M and has 2.25*10^-2M NaOH added?
Tuesday, May 8, 2012 at 4:51pm by Pippi

Chemistry
Consider the titration of 100.0 mL of 0.260 M propanoic acid (Ka = 1.310−5) with 0.130 M KOH. Calculate the pH of the resulting solution after each of the following volumes of KOH has been added. (Assume that all solutions are at 25°C.) (a) 0.0 mL (b) 50.0 mL (c) 100.0 ...
Wednesday, March 13, 2013 at 2:56am by Margaret

chemistry
A 125.0 mg sample of an unknown, monoprotic acid was dissolved in 100.0 mL of distilled water and titrated with a 0.050 M solution of NaOH. The pH of the solution was monitored throughout the titration, and the following data were collected. Determine the Ka of the acid. ...
Tuesday, March 11, 2014 at 3:22pm by Mohanad

chem
the pH of a .100 M solution of sodium formate = 8.37. Calculate the value of Ka of formic acid. okay so I know that the anion of a weak acid is itself a weak base so A- +H2O---> OH- + HA pOH =14-8.37 so conc of OH is 10^-5.63 Ka=[OH][HA]/[A-] so we know the conc of OH and A...
Wednesday, May 13, 2009 at 10:07pm by Natash

bio chem
Explain how to prepare 1L of a buffer at pH 7.4 using 1M hypochlorous acid (Ka = 3.5 x 10-8) and a 1 M NaOH solution. Indicate the volume of hypochlorous acid and sodium hydroxide to use. pH = pKa + log ([A-]/[HA]) Ka= [H+][A-]/[HA] Molarity = moles/L moles = g/molecular weight
Tuesday, March 12, 2013 at 10:31am by Anonymous

AP Chemistry
(pure CH3COOH) Ka = 1.8 x 10-5 sodium acetate NaCH3COO CH3COOH --> CH3COO + H Ka= [H+][CH3COO-]/[CH3COOH] = 1.8 x10^-5 [H+]= 4.85 pH = pKa + log [base]/[acid] 5.0 (chosen pH of buffer) = 4.85 + log [base]/[acid] .15 = log [base]/[acid] 10^0.15 = [1.41 M]/[1 M]
Monday, January 3, 2011 at 10:20pm by Ana

chemistry
Methanoic acid HCO2H(aq) also known as formic acid, is partly responsible for the characterisitic itchy rash produced by the leaves of the stinging nettle plant. Calculate the pH of 0.150 mol/L methanoic acid. The Ka for methanoic acid is 1.8 x 10^-4. My Work: Lets call ...
Sunday, April 6, 2008 at 10:47pm by Sarah

chemistry
Lactic acid, HC3H5O3(aq) is a weak acid that gives yougurt its sour taste(Yeeeeecccckkk). Calculate the pH of a 0.0010 mol/L solution of Lactic acid. The Ka for lactic acid is 1.4 x 10^-4 For Further Reading chemistry - DrBob222, Saturday, April 5, 2008 at 7:39pm Let's call ...
Sunday, April 6, 2008 at 10:42pm by Sarah

physics
Four charges −8 × 10^−9 C at (0 m, 0 m), −7 × 10^−9 C at (5 m, 1 m), −1 × 10^−9 C at (−2 m, −3 m), and 2 × 10^−9 C at (−2 m, 4 m), are arranged in the (x, y) plane as shown. Find the magnitude of the resulting force ...
Wednesday, January 29, 2014 at 11:26pm by sarah

chemistry
I have no clue, please help and explain! =) Find the pH of a 0.150 M solution of a weak monoprotic acid having Ka=1.2×10−5. Find the percent dissociation of this solution. Find the pH of a 0.150 M solution of a weak monoprotic acid having Ka=1.4×10−3. Find the ...
Monday, December 6, 2010 at 7:33pm by Lisa, Help needed!

Chemsitry
A chemist needs to make 50 mL buffer solution with a pH value of 7.25 using the acid/conjugate base pair of KH2PO4 and Na2HPO4 (which has an effective pKa = 6.86). The buffer solution needs to be made so that the total concentration of buffer is 0.10 M (that is, [acid] + [base...
Monday, April 2, 2012 at 11:39pm by Cassandra

Science
Describe the procedure for preparing 250mL of buffer solution with a pH of 5.00 from acetic acid (CH3COOH) and sodium acetate. How many grams of acid do you have to mix with 3.6 grams of sodium acetate? (Given Ka for acetic acid = 1.8x10^-5)
Monday, October 25, 2010 at 12:29am by Chemisty-Buffers

chemistry
There are a number of materials that can be used for neutralizing acid spills. The most common household material is sodium bicarbonate, NaHCO3 (baking soda). sodium bicarbonate (now the IUPAC says we should call it sodium hydrogen carbonate) plus acid (substitute a name here...
Wednesday, November 3, 2010 at 5:26pm by DrBob222

Chemistry
Write a net ionic equation and calculate K for the reaction in aq between acidic acid and sodium hydroxide. Given Ka (acidic acid) = 1.8 x 10^-5
Monday, April 4, 2011 at 8:44pm by Paul

Chemistry
You're right. HCl is the strongest acid. Note that all of the other acids are weak and have a Ka. Look in your text for Ka for each. Then rank the Ka values from weak to strong. The smallest Ka will be the weakest acid. #2. All of these are salts and K^+ is not hydrolyzed. The...
Sunday, October 23, 2011 at 12:07pm by DrBob222

chemisty
0.100 M solution of a weak acid, HX, is known to be 15% ionized. The weak acid has a molar mass of 72 g/mol. 1. What is Ka for the weak acid? 2. What is the pH of the buffer prepared by adding 10.0 g of the sodium salt of the acid (NaX) to 100.0 mL of 0.250 M HX
Wednesday, April 4, 2012 at 5:35am by Clayton

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