Number of results: 10,953
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 ...
Monday, April 9, 2012 at 8:59pm by Sam
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
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 ...
Tuesday, July 27, 2010 at 9:45pm by rick
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
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...
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
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, #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
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 ...
Tuesday, March 19, 2013 at 8:22pm by Devron
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 ...
Thursday, February 12, 2009 at 7:46pm by DrBob222
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
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
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 ...
Wednesday, May 26, 2010 at 1:31am by michael
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 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 ...
Tuesday, October 27, 2009 at 7:32pm by Anonymous
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
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
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
chemsitry
buffer calculate the ph of the .1m propanoic acid Ka 1.35e-5
Sunday, April 10, 2011 at 9:09pm by neee
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
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
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 %...
Tuesday, May 22, 2012 at 8:40am by DrBob222
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
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...
Sunday, October 23, 2011 at 12:07pm by DrBob222
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
Trichloroacetic acid (CCl3CO2H) is a corrosive acid that is used to precipitate proteins. The pH of a 0.050 M solution of trichloroacetic acid is the same as the pH of a 0.040 M HClO4 solution. Calculate Ka for trichloroacetic acid.
Wednesday, November 23, 2011 at 7:10pm by Kaleen
chemistry
Is that 0.050 M acetic acid? Let HAc stand for CH3COOH. You want pH = 3.80, convert that to (H^+) by pH = -log(H^+). Ka = (H^+)(Ac^-)/(HAc) Substitute (H^+),(Ac^-), and Ka into the Ka expression and solve for (HAc). Then substitute into the dilution equation to finish. m1C1 = ...
Friday, December 2, 2011 at 10:04pm by DrBob222
Chemistry
.050 mole of weak acid (HA) is dissolved in enough water to make 1.0 L of solution. The pH of this solution is 3.50. What is the value of Ka for this weak acid?
Friday, April 6, 2012 at 5:52pm by Hannah
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
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
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
You have the Ka for the weak acid is Ka = (H^+)(A^-)/(HA) If we solve this for (H^+) we get ((H^+) = Ka*(HA)/(A^-) At the exact half-way mark to the equivalence point, the acid that is left (not yet neutralized) exactly equals the salt formed; therefore, (HA) = (A^-). Thus, (H...
Friday, March 12, 2010 at 7:28pm by DrBob222
chemistry
Let's call acetate, Ac^- just to save some typing. Ac^- + HOH ==> HAc (acetic acid) + OH^- Kb for acetate = (Kw/Ka) where Ka is the acid constant for acetic acid. Kb = (Kw/Ka) = (HAc)(OH^-)/(Ac^-) Let HAc = x = OH, then (Kw/Ka) = x^2/Ac^-. Kw you know. Ka you know. ...
Monday, November 15, 2010 at 12:20am by DrBob222
Chemistry
Let's call trichloroacetic acid HT and perchloric acid is HClO4. .............HClO4 ==> H^+ + ClO4^- initial.......0.04.....0......0 change......-0.04......0.04...0.04 equil..........0.......0.04....0.04 ............HT ==> H^+ + T^- initial.....0.05...0.........
Wednesday, November 23, 2011 at 7:10pm by DrBob222
Chemistry
In a 1.760 M aqueous solution of a monoprotic acid, 3.21% of the acid is ionized. What is the value of it's Ka? x=1.760M * 3.21/100 - .0565M Ka expression is Ka= {[H+][A-]} / [HA]
Sunday, March 18, 2012 at 10:58pm by Jamie
chemistry
1. if 15.0mL of 4.5 M NaOH are diluted with water to a volume of 500mL, what is the molarity of the resulting solution? 2. in a acid-base titration, 33.65mL of an 0.148 M HCL solution were required to neutralize 25.00mL of a NaOH solution. What is the molarity of the NaOH ...
Tuesday, October 25, 2011 at 10:55pm by djella
Chemistry
Let's call propanoic acid HP. .........HP ==> H^+ + P^- initial.0.3M.....0.....0 change...-x.......x.....x equil....0.3-x....x......x Ka = (H^+)(P^-)/(HP) Substitute from the ICE chart above and solve for x = (H^+). %ion = [(H^+)/(HP)]*100 = [(?H^+)/0.3]*100 Post ...
Monday, April 9, 2012 at 8:59pm by DrBob222
CHEMISTRY LAB urgent assistance
Start with the acid dissociation equation: HAc -> H+ + Ac- Ka is then [H+][Ac-]/[HAc] {given in the question} at the start HAc -> H+ + Ac- and we start from A mole/L and B mol/L for the acid and buffer A M ___0__ B M at the end HAc -> H+ + Ac- A-x____x___B...
Thursday, September 22, 2011 at 7:33am by Dr Russ
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
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
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
Chem.
If you are talking about CH3CH2COOH where COOH is the acid part and R is everything else, then R | C=O | O | H then that is propanoic acid.
Sunday, June 7, 2009 at 8:48pm by DrBob222
Chemistry
We will call weak acid A as HA. pH = 2.20; therefore, H^+ = 6.31E-3 ...........HA ==-> H^+ + A^- I........0.0870.....0.....0 C...........-x.......x.....x E......0.0870-x.....x......x You know x = 6.31E-3; therefore, Ka = (H^+)(A^-)/(HA) Ka = (6.31E-3)(6.31E-3)/(0.08069...
Friday, April 12, 2013 at 12:46pm by DBob222
Chemistry Acid Question Ka
Enough of a monoprotic acid is dissolved in water to produce a 0.0165 M solution. The pH of the resulting solution is 2.38. Calculate the Ka for the acid.
Sunday, April 22, 2012 at 10:52pm by RZeal
ap chemistry
Propanoic acid = HPr HPr ==> H^+ + Pr^- Set up an ICE chart and substitute into the below equation. Ka = (H^+)Pr^-)/(HPr) Solve for (H^+) and convert to pH. (Pr^-) = (H^+) You can find OH^- from (H^+)(OH^-) = 1E-14 %diss = (H^+)/(HPr)*100 = ??
Wednesday, April 6, 2011 at 8:16pm by DrBob222
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...
Sunday, March 15, 2009 at 3:35pm by Rushi
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...
Sunday, March 15, 2009 at 3:34pm by Saira
chemistry
Not quite. Let's call propionic acid (Propanoic acid) HPr to avoid all that typing. HPr ==> H^+ + Pr^- K = (H^+)(Pr^-)/(HPr) Now, we have 37 g HPr/L which is 37/74 = 0.50 mole/L = 0.50 M. Set up an ICE chart. (H^+) = x (Pr^-) = x (HPr) = 0.5-x Substitute into Ka ...
Tuesday, July 27, 2010 at 9:45pm by DrBob222
Chemistry
I mean how it effects the acid-dissociation constant(Ka). How does water effect the Ka value of a weak acid?
Monday, October 27, 2008 at 11:45am by Jessica
chemistry
The acid is acetic acid. Either in your problem or in your text the Ka should be listed as close to 1.8E-5 which is a pKa of 4.74 (pKa = -log Ka). Use the value in your text or notes, not the one I used. The base is acetate. The acid is acetic acid in the HH equation.
Sunday, December 5, 2010 at 8:58pm by DrBob222
Chemistry
Known: Cyanaocetic acid Ka=3.55*10^-3 ionization equation: HC3H2NO2<==> H^+ + C3H2NO2^- Unknown: pH of 0.4M of Cyanaocetic acid I can't figure out how to calculate this. I only knew how to calculate the pH value of an acid with a known Ka, but not in this ...
Monday, April 20, 2009 at 11:42pm by anonymous
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
pKa is the pKa for acetic acid. Since Ka for acetic acid is about 1.8 x 10^-5, the pKa is -log Ka = about 4.75 or so. You need to look up Ka for acetic acid in your text or notes and take the negative log of that value.
Sunday, July 19, 2009 at 7:24am by DrBob222
Chemistry
You must recognize that this is a mixture of two weak acids; i.e., formic acid and acetic acid. I looked up Ka for both and used 1.77E-4 for Ka HCOOH and 1.8E-5 for Ka CH3COOH. Calculate the H^+ from the strong acid, then add the H^+ from the weak acid. Formic acid first since...
Sunday, February 26, 2012 at 2:27am by DrBob222
Chem
SrSO4, CuCN, and Ni(OH)2 are more soluble in acidic solution than in pure water. If you don't understand what basic anions are, then look at it this way. Which anions will hydrolyze with H2O to form a WEAK acid when added to H^+; i.e., an acid that has a Ka (not a strong ...
Thursday, May 2, 2013 at 7:28pm by DrBob222
chemistry
Based on what I can find on the internet, ascorbic acid is a monoprotic acid. If that is so, then let's call ascorbic acid HC. Then HC ==> H^+ + C^- Ka = ((H^+)(C^-)/(HC) If pH = 2.40, then (H^+) = 0.00398/ (C^-) is the same. (HC) = 0.2 = 0.00398 Plug in the Ka ...
Sunday, April 6, 2008 at 12:46am by DrBob222
chemistry
Weak acid is HA. HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) Convert pH to H^+, substitute into Ka expression for Ka and solve for Ka.
Friday, September 24, 2010 at 4:43pm by DrBob222
chemistry
Hey - just look up the Ka (or Kb respectively) values for each compound, and see which is bigger. Stronger acid = bigger Ka and that will dictate which direction it will go in. In this case, H3PO4 is the strongest acid (highest Ka) and therefore the direction of that reaction ...
Sunday, March 8, 2009 at 1:58am by Julia
chemistry
A 0.25 mol/L solution of benzoic acid, KC7H5O2 and antiseptic also used as a food preservative, has a pH of 2.40. Calculate the Ka of benzoic acid at SATP My work: C6H5COOH --> C6H5COO^- + H^+ Ka = (C6H5COO-)(H+)/ C6H5COOH I know my next steps are to use pH = log(H+) to...
Monday, April 7, 2008 at 5:15pm by Dustin
Chemistry
I will do the third one. NH4OH: .010l*.3=.0030 moles HCl: .030*.1=.003 moles they are all react, no excess acid or base. So, pH=7 the fourth one: moles base: .0030 moles acid: .040*.1=.004, or an excess of acid of .001moles volume+ 10ml+40ml=.050 liters concen acid: .001/.050...
Saturday, March 30, 2013 at 7:35pm by bobpursley
Chemistry Very Urgent!!
A 0.035 M solution of a weak acid (HA) has a pH of 4.88. What is the Ka of the acid? HA ==>H^+ + A^- Ka = (H^+)(A^-)/(HA) pH = -log(H^+) You know pH. Convert that to (H^+). (H^+)=(A^-) so plug those into the expression for Ka. (HA) = 0.035 M - (H^+). Plug that in. ...
Monday, April 2, 2007 at 11:35pm by Paul
Chemistry
Let's just call this acid HA. It's the salt that is hydrolyzing, so A^- + HOH ==> HA + OH^- Kb = Kw/Ka = (HA)(OH^-)/(A^-) You know Kw, Ka is what you want, (HA)=(OH^-) and you can get the OH from the pH. After you find Ka, use that as you would a weak acid ...
Thursday, January 21, 2010 at 4:24pm by DrBob222
Chemistry
Propanoic acid is CH3CH2COOH. Let's simplify that by calling it HPr. HPr + NaOH ==> NaPr + H2O I have no idea what an IRE-C table is. moles NaOH = L x M = 0.01350 x 0.200 M = 0.027 moles HPr = L x M = 0.05000 x 0.100 = 0.005. initial concns: HPr = 0.005 moles/0.0635...
Monday, April 26, 2010 at 7:54am 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
An acid HA has Ka = 2.28 x 10^-4. The % ionisation of this acid in a 0.170 M solution of the acid in water is? An acid HA has Ka = 2.28 x 10^-4. The % ionisation of this acid in a 0.170 M solution of the acid in water is closest to a. 2.930 % b. 3.662 % c. 2.564 % d. 0.498 % e...
Thursday, May 26, 2011 at 9:21pm by Cheryl
chemistry
Weak Acid 1.) The pH of a 0.060M weak monoprotic acid HA is 3.44. Calculate the Ka of the acid. 2.) The pH of 0.100M solution of weak monoprotic acid HA is 2.85. What is the Ka of the acid?
Thursday, January 20, 2011 at 5:19am by jaycab
chemistry
Weak Acid 1.) The pH of a 0.060M weak monoprotic acid HA is 3.44. Calculate the Ka of the acid. 2.) The pH of 0.100M solution of weak monoprotic acid HA is 2.85. What is the Ka of the acid?
Thursday, January 20, 2011 at 5:19am by jaycab
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
A. Strong Base 1.) What is the concentration of a solution of KOH for which the pH is 11.89? 2.) What is the pH of a 0.011M solution of Ca(OH)2? B. Weak Acid 1.) The pH of a 0.060M weak monoprotic acid HA is 3.44. Calculate the Ka of the acid. 2.) The pH of 0.100M solution of ...
Thursday, January 20, 2011 at 3:37am by jaycab
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...
Wednesday, March 13, 2013 at 2:56am by Margaret
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 ...
Sunday, April 6, 2008 at 10:42pm by Sarah
Chem
Acetic acid is 3% dissociated in a 0.02M solution. Give the concentrations of each of the two ions, the acetic acid, and the Ka for acetic acid. HC2H3O2 ==> H^+ + C2H3O2^- So (H^+) = 0.02*0.003 = ?? (C2H3O2^-) = 0.02*0.003 = ?? Plug into Ka expression and solve for Ka. ...
Saturday, July 14, 2007 at 2:20pm by Lizette
Chemistry
Given acidity constant,Ka of benzoic acid is 6.28 X 10^-5. pH of 0.15 molar solution of this acid is ? Here is my method: Benzoic acid is a weak acid,hence it dissociates very little. So: C6H5COOH---> C6H5COO- + H+ [H+] and [C6H5COO-] are yet to be calculated,so let ...
Friday, March 23, 2012 at 12:28am by Meenaakshi
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...
Wednesday, February 27, 2008 at 4:22pm by Sarah
chemistry
Two comments. #1. You don't means nitric acid. You mean nitrous acid, HNO2. #2. You found the Ka for HOCl, HNO2, and C6H5COOH. You want Kb (not Ka) for the anion; i.e., OCl^-, NO2^-, and C6H5COO^-. Easy to do. KaKb = Kw. You know Kw, you find Ka in the tables for the ...
Tuesday, April 8, 2008 at 8:15pm by DrBob222
chem
You COULD work each one for pH and see which is the lowest but that isn't necessary. I'll show you how to do one of them, HF since it is first. .............HF ==> H^+ + F^- initial......1.0.....0.....0 change.......-x......x......x equil......1.0-x.....x..........
Friday, January 20, 2012 at 4:00pm by DrBob222
chemistry
Look in your text or notes at the Ka for weak acids. If a Ka is listed it is a weak acid. If not it is a strong acid (or an incomplete table.
Wednesday, April 28, 2010 at 2:52pm by DrBob222
Chem 2
What is the approximate value of the K for the neutralization of nitrous acid with ammonia? Ka for the acid is 0.00045 and Kb for the base is 0.000018. Do I multiply Ka*Kb?
Thursday, February 28, 2008 at 10:30pm by Sarah
Chemistry
The Ka for a particular weak acid is 4.0 x 10-9. Calculate the pH of a 0.040 M solution of this acid. whats the relation b/w Ka and pH?
Saturday, November 1, 2008 at 2:51am by A.A
Chemistry
What is the lewis structure for Propanoic Acid?
Saturday, December 4, 2010 at 5:49pm by Eddie
Chemistry
An 0.0284 aqueous solution of lactic acid is found to be 6.7% ionized. Determine Ka for lactic acid. HC3H5O3 + H2O <-- --> H3O^+ + C3H5O3^- Ka =? Thanks.
Monday, August 10, 2009 at 12:46am by Sev
Chemistry
(a) Explain what is meant by a weak acid. Methanoic acid (HCO2H) has a numerical value of Ka(T) at 298K of 1.60 × 10-4 (b) Write an expression for Ka(T) of methanoic acid, and state its units. (c) What is the pH value of a 0.10 mol dm-3 aqueous solution of methanoic acid...
Sunday, April 22, 2012 at 3:44pm by Henry
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
College Chem II
Benzoic acid = HBz HBz ==> H^+ + Bz^- Ka = (H^+)(Bz^-)/(HBz) Set up an ICE chart, substitute into the Ka expression, and solve for (H^+), then convert to pH. You will need to look up the Ka for benzoic acid.
Sunday, March 7, 2010 at 6:07pm by DrBob222
chemistry
Reading the pH at the 1/2 way point to the equivalence oint will give you the Ka if you set pH = -log(H^+) Reading the pH will give you the pKa of the acid directly. Try this. For an acid HA that ionizes HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) at the half way point, (A...
Tuesday, April 8, 2008 at 11:39pm by DrBob222
Chem 2
A 0.15 M solution of butanoic acid, C3H7COOH, has a pH of 2.8210. Find the Ka of butanoic acid. Call buanoic acid HB. Then the ionization is HB ==> H^+ + B^- To start (HB) = 0.15 (H^+) = 0 and (B^-) = 0 After ionization, (H^+) = x but that is found by converting pH to (...
Thursday, July 19, 2007 at 1:47am by Jess
Chemistry
I believe you want propanoic acid and isobutyl alcohol.
Monday, October 24, 2011 at 1:22pm by DrBob222
college
Acetylene is an alkyne, benzene is an aromatic as is toluene, xylene, phenol, etc. Ethyl ether, dimethyl ether, dipropyl ether, etc. Acids are acetic acid, propanoic acid, benzoic acid, etc. I don't think you will get any takers on the harmful effects. BOOKS of several ...
Sunday, October 24, 2010 at 5:12pm by DrBob222
Chemistry
Calculate the pH of a solution that is 0.060 M in potassium propionate(C2H5COOK or KC3H5O2) and 0.085 M in propionic acid (C2H5COOH or HC3H5O2)
Thursday, October 28, 2010 at 1:47pm by Anonymous
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 ...
Monday, April 8, 2013 at 7:01pm by Marcus
Chemistry
The pH at the halfway point of a weak acid/srong base or weak base/strong acid is pH = pKa. Look at the Ka expression for a weak acid. Ka = (H^+)(A^-)/(HA). Now solve this for H^+. (H^+) = Ka*[(HA)/(A^-)]. When you are halfway there, the (HA) = (A^-); therefore, Ka = (H^+). If...
Saturday, December 1, 2012 at 3:03pm by DrBob222
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
chem
Let's call sorbic acid HSorb. HSorb ==> H^+ _ Sorb^- Ka = (H^+)(Sorb^-)/(HSorb) = Ka. So (H^+) = x (Sorb^-) = x (HSorb) = 0.37 - x Substitute into the Ka expression and solve for x. Note: It would be a good idea to convert pKa to Ka first.
Sunday, February 21, 2010 at 5:22pm by DrBob222
chem 12
You start over. Benzoic acid is an acid, not a base. Let's call benzoic acid HB. ..........HB ==> H^+ + B^- init.....0.15.....0.....0 change....-x......x.....x equil....0.15-x...x.....x Ka + (H^+)(B^-)/(HB) Substitute the equilibrium line above into the Ka ...
Tuesday, October 11, 2011 at 2:03pm by DrBob222
chemistry
Yes to the first question. (H^+) = 10^-pH. At the half way point, which is what you had, then Ka = (H^+ or pKa = pH but that is only true at the half way point. You can see for a weak acid, such as HA HA ==> H^+ + A^- and Ka(H^+)(A^-)/(HA). Solving for (H^+) we get (H...
Thursday, April 17, 2008 at 3:04pm by DrBob222
chemistry
Calculate [acetate] of 0.057 mol/L solution of acetic acid of Ka = 1.76x10^-5 i know to find Ka it's: Ka= products/reactants so i tried doing an ICE table with the equation to be: CH3CO2 + H2O <--> CH3COOH + H am i on the right track? if not...help please!
Wednesday, July 28, 2010 at 12:06am by Kat
CHEM HELP!
Nicotinic acid (niacin) is a monoprotic acid with the formula HC6H4NO2. A solution that is 0.012M in nicotinic acid has a pH of 3.39 at 25*C. What is the acid-ionization constant, Ka and pKa for this acid at 25*C?
Tuesday, March 17, 2009 at 6:36pm by Ashley
College Chemistry
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 3.30 at 25C. what is the acid-ionization constant, Ka and pKa for this acid at 25C?
Thursday, March 5, 2009 at 2:35pm by Anonymous
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