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April 16, 2014

Search: Chlorous acid, HClO2, has a Ka = 1.1 x 10-2. Determine the pH of a 0.27 M solution of HClO2. 0.28 0.57 1.14 1.26 1.31

Number of results: 96,844

chemistry
What is the percent ionization of 0.025 M chlorous acid, HClO2 solution? Ka HClO2 = 1.1*10^(-2) Answer: 48% What I did: HClO2(aq) + H2O(l) -><- ClO2(aq) + H3O(aq) initial-change-end table results: x^(2)/(.025-x)=1.1*10^(-2) x=1.6*10^(-2) % ionization = (1.1*10^(-2))/(1.6...
Saturday, July 30, 2011 at 7:13pm by % ionization

Chemistry
A buffer was prepared by dissolving 22.61g of NaCLO2 into 0.100L of 4.00M aqueous HCLO2. Calculate the pH of the original Chlorous acid/chlorite ion buffer. Ka= 1.1 x 10^-2.
Monday, December 3, 2012 at 10:14am by May

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)hydrobromic acid (HBr) b)chlorous acid (HClO2) c)benzoic acid (C6H5COOH)
Wednesday, February 29, 2012 at 10:35pm by DANNY

chem
Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 0.080 M NaOH. chlorous acid (HClO2) benzoic acid (C6H5COOH)
Saturday, March 23, 2013 at 6:28pm by hannah

Chemistry
HClO2 is a weak acid. HClO2 + H2O ==> H3O^+ + ClO2^- Ka = (H3O^+)(ClO2^-)/(HClO2) Set up an ICE chart, calculate (H^+) and convert that to pH.
Monday, March 8, 2010 at 8:22pm by DrBob222

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

12th grade
The series of H, Cl, and O acids are as follows: HClO4 = perchloric acid HClO3 = chloric acid HClO2 = chlorous acid HClO = hypochlorous acid The general rule is that the acid is stronger if it has more O atoms in a series such as this. HClO4, perchloric acid, is a very strong ...
Tuesday, July 27, 2010 at 10:29pm by DrBob222

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--To Rose
HClO is hypochlorous acid, The question was for chlorous acid, HClO2.
Wednesday, November 9, 2011 at 8:25pm by DrBob222

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
How do you name this acid? HClO2 I found the answer but now I need to know N2F4. chlorous acid dinitrogen tetrafluoride
Saturday, December 16, 2006 at 4:08pm by Carter

Chemistry -- Please Help!!
Find the pH of a mixture of .150 M HF(aq) solution and 0.100 M HClO2(aq) HF + H2O <--> F- + H3O+ Ka= 3.5*10^-4 So I will show my attempt below... [HF] [F-] [H3O+] I .150 0 0 C -x +x +x E .15-x x x Ka= [F-][H3O+]/[HF] 3.5*10^-4= x^2/(.150-x) assume x is small so (.150-x...
Monday, April 26, 2010 at 7:06pm by Erin

Chm 2
The question asks, Consider the following weak acids and their Ka values Acetic acid Ka = 1.8x10^-5 Phosphoric acid Ka = 7.5x10^-3 Hypochlorous acid Ka = 3.5x10^-8 What weak acid-conjugate base buffer system from the acids listed is the best chouce to prepare the following ...
Tuesday, October 15, 2013 at 12:29am by Kyle

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

chemistry
a) carbon monoxide. b) no such animal. You may mean HClO2 which is chlorous acid.
Tuesday, February 8, 2011 at 2:55pm by DrBob222

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

chemistry
Can someone please explain how to do this question? A 0.057M solution of a weak acid has a pH of 4.93. Determine the ionization constant, Ka, of the acid. a) 2.1 x 10^-4 b) 2.4 x 10^-9 c) 4.1 x 10^-8 d) 2.8 x 10^-7 e) 1.4 x 10^-10 -- Well, to be honest, I skipped this. I know ...
Thursday, April 7, 2011 at 10:18pm by polkadot

Chemistry
A) Determine the pH of a 0.98 x 10^-2 mol L solution of hydrocyanic acid (HCN) Ka = 4.0 x 10^-10 B) What is the pH of a 0.243mol L solution of methylamine? (pKb for CH3NH2 = 3.30) C) determine the pH of a buffer solution of Na2CO3 (pkb = 3.68, 0.125 M) and NaHco3 (0.35M)
Monday, June 3, 2013 at 10:07pm by Jake

Chem
Aspirin (acetylsalicylic acid, ) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolved 2.00 g of aspirin in 0.600 L of water and measured the pH . What was the Ka value calculated by the student if the of the pH of the solution was 2.62? I...
Sunday, March 28, 2010 at 8:03pm by Katie

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
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

chem
NaOH + HClO2 ==> NaClO2 + H2O so the pH at the equivalence point is determined by the hydrolysis of NaClO2 (actually the hydrolysis of the ClO2^-. The (ClO2^-) = 0.0444M. If you don't know how to do this I can show you. .........ClO2^- + HOH ==>HClO2 + OH^- I........0....
Saturday, March 23, 2013 at 6:28pm by DrBob222

Chemistry
You might want to look up Kb or an amide vs Kb for NH3. NH3 is the stronger base. Compare Kb for each. HClO3 and HClO2 are acids. HClO3 is the stronger acid; therefore, HClO2 is the stronger base. Compare Ka values. HClO2 has a Ka, HClO3 is a strong acid. Cl3CCOOH is a ...
Tuesday, November 20, 2007 at 1:51am by DrBob222

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
........HClO2 + H2O ==> H3O^+ + ClO2^- I.....0.120..............0.......0 C........-x..............x.......x E.....0.120-x............x.......x Substitute the E line into the Ka expression for HClO2 and solve for x = (H^+), then convert to pH = -log(H^+)
Saturday, February 9, 2013 at 11:00pm by DrBob222

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 them be...
Friday, March 23, 2012 at 12:28am by Meenaakshi

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
[HNO2] = .0410 M [NaNO2] = .0250 mol / .250 L = .100 M [H+] = Ka x [HA (weak acid)] / [A- (conj. base)] Ka = 6.0 x 10^-4 (a given constant) [H+] = (6.0 x 10^-4) x [.041]/[.100] = 2.5 x 10^-4 M. pH = -Log[H+] -- pH = -log[2.5 x 10^-4] pH = 3.61
Sunday, April 29, 2007 at 7:30pm by Josh

Chemistry
What is the percent ionization for each of the following acids? a. 0.022 M HClO2 solution of pH= 3.88 Answer obtained: .60% pH = 10^(-3.88) = 1.32 x 10^-4 % ionization= [H+] formed divided by MHA (original acid concent) x 100 = (1.32x10^-4)/0.022 x 100 = .60% b. 0.0027 M HClO2...
Sunday, April 3, 2011 at 11:11pm by Cassie

Chemistry
I apologize, HAc + H20---> H30+ Ac HAc= acetic acid, which is a weak acid. a.) Ka=1.8 x 10–5=[x][x]/0.100M-x 5% rule allows you to ignore -x Ka=1.8 x 10–5=x^2/0.100M solving for x, x=1.34 x 10^-3 pH=-log(1.34 x 10^-3 M) B.) pKa=-log(Ka) pH=pKa+log([Ac/HAc]) Concentrations ...
Tuesday, February 26, 2013 at 8:00pm by Devron

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 case. Also, I ...
Monday, April 20, 2009 at 11:42pm by anonymous

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

General Chem
A weak acid has Ka= 1*10^-3. If [HA]= 1.00 M what must be [A-] for the pH to be pH 2.7? My work: pH= Pka + log(B/A) 2.7= 3 + log (B/A) Kw= Ka* Kb 1*10^-14= 1*10^-3 * Kb Kb= 1*10^-11 I don't know how to go about from here
Sunday, August 9, 2009 at 10:02pm by Hina

chem
Ice table ? I haven't heard of that term in awhile. Well since acetic acid is a weak acid and NaOH is a strong base... 120ml 0.15M acetic acid 30ml 0.2M NaOH -You need Ka for the weak acid. (for acetic acid Ka= 1.8x10^-5) steps: 1. Determine if moles H+ or moles OH- is in ...
Monday, August 27, 2007 at 10:42pm by ~christina~

chemistry
never mind, I found out how to do this =) (the post above was me by the way). set up I.C.E. table for the equilibrium expression. Since we know that Ka value for acetic acid is about 1.76 x 10^-5, we can solve for x (which is [H+] in this case). Ka = x^2 / 0.10 - x => ...
Sunday, May 4, 2008 at 5:45pm by Anonymous

Chemistry
Henderson-Hasselbalch equation: pH=pKa+log((base)/(acid)) Kb*Ka=1.0*10^-14 1.74*10^-5*Ka=1.0*10^-14 Ka=1.0*10^-14/1.74*10^-5 Ka=5.75*10^-10 pKa=-log(5.75*10^-10) pKa=9.24 pH=9.24+log((.246)/(.0954)) pH=9.65
Sunday, April 15, 2012 at 6:55am by Whitney

Chemistry
Part A: Unknown Acid use 1gram and mix with 120 mL of distilled water Determine the concentration of the acid by titrating with 0.0998 M of NaOH Concentration of NaOH= 0.0998M Volume of NaOH= 3.5 mL # moles of NaOH= Initial Concentration of Weak Acid- Show Calculation of Ka ...
Wednesday, March 24, 2010 at 3:40pm by Saira

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. Calculate...
Monday, April 2, 2007 at 11:35pm by Paul

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

General Chem
weak acid has Ka= 1*10^-3. If [HA]= 1.00 M what must be [A-] for the pH to be pH 2.7? My work: pH= Pka + log(B/A) 2.7= 3 + log (B/A) You are OK to here. You need not use any of what follows. The problem gives you (HA)= 1.00 M [the acid in the problem) and it asks you to solve ...
Sunday, August 9, 2009 at 10:02pm 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.......x Look up Ka...
Friday, January 20, 2012 at 4:00pm by DrBob222

chemistry
2- A 0.310 M solution of a weak acid, HX, has a pH of 2.53 a. Find the [H+] and the percent ionization of nitrous acid in this solution. b. Write the equilibrium expression and calculate the value of Ka for the weak acid. c. Calculate the pH of the solution formed by adding 10...
Sunday, March 25, 2012 at 6:06pm by bob

chem
1. Determine if moles H+ or moles OH- is in excess 2. Determine ammount of excess. 3. Determine Molarity of excess ammount. 4. Solve for pH directly or by pOH id OH- is in excess I copied your original post above. I won't argue with you Christina for only you know what you ...
Monday, August 27, 2007 at 10:42pm by DrBob222

chemistry
The Ka of hypochlorous acid (HClO) is 3.0 × 10-8 at 25°C. Calculate the pH of a 0.0385-M hypochlorous acid solution? i got help with this problem, but i dont get where one of the numbers came from (3.4 x 10^-5)??? i tried makeing 3.0 x 10^-8 the x but i still dont get that ...
Tuesday, November 23, 2010 at 2:38pm by jessica

Chem202 - for DrBob222
I'm afraid I can't be much help. 1. I worked this from scratch tonight and I also come out with pOH = 6.70 and pH = 7.30 (with or without the quadratic). 2. I searched the old files and found a number of posts I had made for this same problem but none of them gave an answer; i...
Sunday, May 26, 2013 at 10:19pm by DrBob222

Chemistry- Acids and Buffers
Cyanic Acid is a weak acid HOCN + H2O <--> H3O+ +OCN- ka= 3.5*10^-4 a) if 2.5ml of 0.01 M cyanic acid solution is added to 25.0 ml of a formis acid buffer with ph = 3.70, what is the ratio of [OCN-]/[HOCN] in the resulting solution? b) determine the pH of a 0.500 M ...
Monday, April 9, 2012 at 4:21am by Tasneem

chem
Calculate the pH of 0.20 M NaCN solution. NaCN ---> Na+ + CN- CN- + H20+ ---> HCN+ + OH- Initial conc. of CN- = 0.20 mol/L change = -x equillibrium = 0.20-x HCN equill. = +x OK to here. OH equill. = 1x10^-7+x This is also +x. I will let you redo the math to get pH but I ...
Friday, July 24, 2009 at 1:07am by DrBob222

Chemistry(Please just check, thank you)
1) The approximate pH of a 3.0 X 10^-3 M solution of the strong acid H2SO4 is? I did pH= -log(3.0 X 10^-3) = 2.5 2)If HCLO2 is a stronger acid than HF, which is stronger than HOCL, then the order of strengths of the conjugate bases of these acids is? My answer is CLO2^-^ < ...
Saturday, March 24, 2012 at 1:50pm by Hannah

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
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 0.010 M solution of a weak monoprotic acid has a pH of 3.70. What is the acid-ionization constant, Ka, for this acid? a. 2.0 ´ 10-4 b. 2.0 ´ 10-5 c. 4.0 ´ 10-6 d. 4.0 ´ 10-7 e. 4.0 ´ 10-8
Thursday, May 13, 2010 at 5:47am by zacky

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
A student prepared a .10M solution of acidic acid. Acidic acid has a Ka of 1.75 x 10-3. What are the hydronium ion concentration and the PH of the solution? I think the PH is 1.76 because I hit -log(1.75 x 10-3) on my calculator and thats what I got thanks in advance for your ...
Wednesday, March 21, 2007 at 6:28pm by Jodi

chem
Calculate the pH of 0.20 M NaCN solution. NaCN ---> Na+ + CN- CN- + H20+ ---> HCN+ + OH- Initial conc. of CN- = 0.20 mol/L change = -x equillibrium = 0.20-x HCN equill. = +x OH equill. = 1x10^-7+x Ka= 6.2 x 10^-10 Kb = KW/Ka = 1x10^-14/6n2 x 10^-10 = 1.6 x 10^-5 1.6 x 10...
Friday, July 24, 2009 at 1:07am by Lena

Chem202 - for DrBob222
Thanks but I'm no guru; Bob Pursley is the all around guru. HClO2 + NaOH ==> NaClO2 + H2O for the neutralization. At the equivalence point the pH is determined by the hydrolysis of the NaClO2 or Kb for ClO2^-. ClO2^- + HOH ==> HClO2 + OH^- is the hydrolysis reaction for ...
Sunday, May 26, 2013 at 10:19pm by DrBob222

Chemistry(Please check answers)
1) The Ka for acetic acid is 1.8e-5. What is the pH of a 3.18M solution of this acid? I did 1.8e-5 = x^2/3.18 x=sqrt 1.8e-5 X 3.18 = 7.56e-3 pH=-log(7.56e-3) = 2.12 The pH is 2.12 2) At 25 degrees celsius, the pH of a 1.75M NaCN solution (the Ka for HCN is 4.0e-10) is ? 4.0e-...
Sunday, April 1, 2012 at 3:40pm by Hannah

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? Give...
Sunday, April 22, 2012 at 3:44pm by Henry

chemistry
Write the benzoic acid ionization. Write the Ka expression. Use pH = =log(H^+) to calculate (H^+). Substitute into Ka expression and determine Ka.
Saturday, April 5, 2008 at 6:42pm by DrBob222

Chemistry/pH- Weak Acid
Hi again! I have a new question, Can you help me? Consider 50.0 mL of a solution of a weak acid HA (Ka = 1.00.E-6), which has a pH of 4.000. What volume of water must be added to make the pH = 5.000? My Calculations: To calculate the concentration of x, I take the pH value -&...
Sunday, August 17, 2008 at 6:07am by Mary

chemistry
You need to know the name of acids. Salts are named for the acid from which the salt is derived. We'll do that first. HCl hydrogen chloride HClO hypochlorous acid HClO2 chlorous acid HClO3 chloric acid HClO4 perchloric acid. For the salts of those acids, the binary is -ide ic ...
Thursday, November 20, 2008 at 11:59am by DrBob222

CHEM
28.42mL of O.1103mol/L KOH titrated / reacted with all the Vit C there was. O.1103mol/L * 28.42mL *1L/1000mL = 3.13*10^-3mol KOH = 3.13*10^-3mol Vit C O.552g/3.13*10^-3mol = ~176g/mol To find Ka use pH=pka + log[base]/[acid] pH is 3.72, Initial [acid]=0.552g/176g/mol = 3.13*10...
Wednesday, March 21, 2012 at 1:30am by Caprice

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+x so Ka is ...
Thursday, September 22, 2011 at 7:33am by Dr Russ

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
What is the pH of an aqueous solution of 0.10 M NaHC(3)H(2)O(4)? Answer: 10.9 What I did: NaHC(3)H(2)O(4) + H(2)O--><-- Na + H(2)C(3)H(3)O(4) + OH H(2)C(3)H(3)O(4) is Malonic acid with a Ka value of 1.5*10^(-3) initial-change-equilibrium (i.c.e) table results: (x^2)/(0....
Saturday, July 30, 2011 at 7:02pm by pH levels

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 ionization. HA...
Thursday, January 21, 2010 at 4:24pm by DrBob222

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

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
At 25 degrees Celsius the Ka for a 0.120 mol/L solution of acetic acid (HC2H3O2 ) is 1.8 X 10-5. determine the pH of the acetic acid solution
Monday, November 15, 2010 at 2:16pm by kneLl

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

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
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(Please check)
1) Which statement below is correct (true)? a) Since HNO2 is a stronger acid than HF it must have a greater pKa. b) Since HOCℓ is a stronger acid than HCN, then OCℓ– is a stronger base than CN–. c) The dissociation (or ionization) of HOCℓ in water is HOC&#...
Saturday, March 24, 2012 at 5:48pm by Hannah

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/pH- Weak Acid
in response to Chemistry/pH- Weak Acid. but the rule of 5%? 0,0100/(1.00x10^-4) =100% The aproximation you do is not valid...or? Ka= (1,0*10^(-4))^2/(x-1*10^(-4) )=1,00*10^(-6)....I got c=0,0101M :-(
Monday, August 18, 2008 at 9:24am by Mary

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/pH- Weak Acid
Hi again! I have a new question, Can you help me? Consider 50.0 mL of a solution of a weak acid HA (Ka = 1.00.E-6), which has a pH of 4.000. What volume of water must be added to make the pH = 5.000? My Calculations: To calculate the concentration of x, I take the pH value -&...
Sunday, August 17, 2008 at 6:01am by Mary

Chemistry
What is the [H+] of a 0.1 M malonic acid with a Ka of 10^-9 solution? I thought it was just (Ka x Ca)^(0.5), but that doesn't get me a pH of 5.
Saturday, May 7, 2011 at 9:34pm by Ed

chemistry
enzoic acid is a weak monoprotic acid with Ka = 6.5×10-5 M. NaOH(s) was gradually added to 1.00 L of 8.66×10-2 M benzoic acid. 1. Calculate the pH of the solution before the addition of the base. 2. Calculate the pH of the solution after the addition of 5.20×10-2 mol of NaOH(s).
Tuesday, March 25, 2014 at 9:14pm by bekah

chemistry
The question asks, "Butanoic acid has a partition coeff of 3.0 when distributed b/t water and benzene. Find the formal concentration of butanoic acid in each phase when 100 mL of 0.10 M aqueous butanoic acid is extracted with 25 mL of benzene at pH 10.00." I know that the ...
Sunday, November 15, 2009 at 8:35pm by Anonymous

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 expression ...
Sunday, April 6, 2008 at 12:46am by DrBob222

chemistry
You don't need to derive the HH equation, only derive the pH range needed. pH = pKa + log(base)/(acid) The problem tells you that acid and conjugate base should not differ by more than a factor of 10. Therefore, use acid = 10*base as the lower end and 10*acid = base as the top...
Tuesday, April 16, 2013 at 11:53am by DrBob222

Chemistry
i. Which of the following solutions has the lowest pH (more acidic). Explain you answers. a. a 0.1 M solution of a strong acid or a 0.1 M solution of a weak acid. b. a 0.1 M solution of an acid with Ka=2×10-3 or one with Ka=8×10-6 c. a 0.1 M solutions of a base with pKb=4.5 or...
Saturday, April 6, 2013 at 3:08pm by m

Chemistry
i. Which of the following solutions has the lowest pH (more acidic). Explain you answers. a. a 0.1 M solution of a strong acid or a 0.1 M solution of a weak acid. b. a 0.1 M solution of an acid with Ka=2×10-3 or one with Ka=8×10-6 c. a 0.1 M solutions of a base with pKb=4.5 or...
Saturday, April 6, 2013 at 9:24pm by m

Chemistry
i. Which of the following solutions has the lowest pH (more acidic). Explain you answers. a. a 0.1 M solution of a strong acid or a 0.1 M solution of a weak acid. b. a 0.1 M solution of an acid with Ka=2×10-3 or one with Ka=8×10-6 c. a 0.1 M solutions of a base with pKb=4.5 or...
Sunday, April 7, 2013 at 7:47am by m

chemistry
What is Ka for 4-aminobenzoic acid if a 0.020 M solution of the acid has a pH of 3.31? a. 2.5 ´ 10-2 b. 2.0 ´ 10-2 c. 4.9 ´ 10-4 d. 1.2 ´ 10-5 e. 2.8 ´ 10-6
Sunday, May 30, 2010 at 2:14pm by james

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-Urgent Help Please
I don't know any tricks for remembering them unless it is to use them often enough that they become second nature. -ide names binary compounds (two atoms to the molecule such as NaCl-sodium chloride or CaC2-calcium carbide). The easiest way to describe the salts is to know ...
Thursday, March 5, 2009 at 9:33pm by DrBob222

chemistry
OK. So you used a computer program to do all this. For the weak acid, which I will cll HA, ionizes HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) Solving this for (H^+), we get (H^+) = Ka*(HA)/(A^-). This is the equation that determines the pH from start to the equivalence point. ...
Friday, May 9, 2008 at 9:47pm by DrBob222

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
Here's what I did: So I set values for: [Acetic Acid] = y [H3O+] = 10^-5 [Acetate] = 0.1M - y Ka = 10^-4.76 And from the equation: Ka = [Acetate][H3O+]/[Acetic Acid], I found y, or the molarity of acetic acid, to be 10^-6/(10^-5 + 10^-7.46). And from this, I also got the ...
Friday, March 18, 2011 at 8:57pm by Sandra

Chemistry
A monoprotic acid with a Ka of 2.92 × 10-5 has a partition coefficient of 4.2 (favoring octanol) when distributed between water and octanol. Find the formal concentration of the acid in each phase when 100 mL of 0.10 M aqueous acid is extracted with 31 mL of octanol at (a) pH ...
Thursday, November 15, 2012 at 9:51pm by Avery

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
Beaker 2 solves for Ka. ...........X^- + HOH ==> HX + OH^- initial...0.07...........0......0 change.....-y............y.....y equil.....0.07-y.........y......y Kb for X^- = (Kw/Ka for HX) = (y)(y)/(0.07-y) Kw, of course, is 1E-14 and you solve for Ka for HX. y = OH and you ...
Sunday, February 12, 2012 at 3:17pm by DrBob222

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
Sacchirin is a monoprotic acid. If the pH of a 1.50x10^-2M solution of this acid is 5.53, what is the Ka of saccharin? (the answer is 5.8x10^-10) I know we're supposed to use the Henderson Hasselbach equation to solve for the [H+] from pH, but what's next? Thanks Dr. Bob.
Tuesday, May 27, 2008 at 4:40pm by howie

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