April 17, 2014

Search: ha

Number of results: 1,161

THats not nice lolhi, wat u say is wat u r, so ha ha ha ha ha ha ha ha
Monday, June 8, 2009 at 10:15pm by ANGRYNESS

4th grade fractions
u need more questions and answers thank u who ever!!!!ha ha ha ha ha ha ha and a million more!!!!!!!!!!!!!!
Wednesday, February 9, 2011 at 3:33pm by poop

oh thank you so much Mrs. Sue, you are the bset. could the others around him say poor guy ha ha ha, and or thats too bad ha ha ha, what do you think the others could say?
Monday, September 21, 2009 at 8:31pm by sara

I am going to assume that you did that correctly. and go from there. ***The problem I believe is that you didn't realize what ionic strength is equal to molarity. Let NaHPO4=A- and let NaH2PO4=HA 2.14=A-/HA and HA +A-=0.1 Solving for A-, A-=0.1-HA Substitute one equation into ...
Tuesday, March 19, 2013 at 7:20pm by Devron

Chemistry Lab
HA + NaOH ==> NaA + H2O mols NaOH = M x L = ? mols HA = mols NaOH (look at the equation. 1 mol HA = 1 mol NaOH) M HA = mols HA/L HA. .........HA ==> H^+ + A^- I........?M......0......0 C.........-x.....x.......x E........?M -x...x.......x Ka = (H^+)(A^-)/(HA) Substitute ...
Tuesday, October 30, 2012 at 1:07pm by DrBob222

ha ha ha ha ha
Saturday, October 13, 2012 at 7:33pm by Just kiding I lied

Ha Ha Ha Ha, you have to submit that today dont you???
Wednesday, July 18, 2012 at 10:34am by Costaatt Student

When one 1 mole of A^- hydrolyzes, it produces 1 mole of HA and 1 mole of OH^-. Or you can make up an ICE chart from your equation. A^- + HOH ==> HA + OH^- initial: A^- = 0.1. HA = 0 OH = 0 change: HA = +x OH^- = +x A^- = -x equilibrium: HA = x OH^- = x A^- = 0.1-x etc. Don...
Wednesday, May 13, 2009 at 10:07pm by DrBob222

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

pH = pKa + log{[A-]/[HA]} or pH = pKa + log{[NaA]/[HA]} if pH = pKa, then {[log{[NaA]/[HA]} = 0 or [NaA] = [HA] The reaction between NaOH and HA is: NaOH + HA --> NaA + H2O Since we start with no NaA, we must add enough NaOH so that [NaA] = [HA] That happens when we ...
Monday, January 12, 2009 at 5:33pm by GK

Do you want the fraction [A-]/[HA] dissociated? [Your value for the pKa for aspirin is low from memory =about 3.5??] From the H-H equation pH=pKa+log([A-]/[HA]) 8.6=5.6 + log([A-]/[HA]) 3.0=log([A-]/[HA]) so 10^+3 = [A-]/[HA]
Monday, June 20, 2011 at 10:30am by Dr Russ

No. H^+ is not equal to A^. HA ==> H^+ + A^- Ka = 3.2 x 10^-4 = (H^+)(A^-)/(HA) The problem tells us that the pH of the stomach is about 1.5 so that is the H^+. 1.5 = -log(H^+) and (H^+) = 0.0316 M. Then (H^+)(A^-)/(HA) = 3.2 x 10^-4 0.0316(A^-)/(HA) = 3.2 x 10^-4 (A^-)/(HA...
Sunday, July 26, 2009 at 9:51pm by DrBob222

No. Think of acetic acid which you have done many times. If we call acetic acid, HA, (and we fan call aspirin HA), then HA <==> H^+ + A^- We know that for every molecule of HA that ionizes, 1 H^+ and `1 A^- are produced but that says nothing about HA and H^+ or HA and A...
Sunday, July 26, 2009 at 9:51pm by DrBob222

Let's call aspirin, HA HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) (H^+) = x (A^-) = x (HA) = 500 mg x 2 tablets = 1000 mg (1 gram and moles is 1g/molar mass with M = mols/0.325L. Solve for x and convert to pH.
Sunday, December 5, 2010 at 8:53pm by DrBob222

For simplicity, let's call cloroacetic acid HA. Then HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) Convert pH to (H^+) from pH = -log(H^+). (A^-) is the same. After equilibrium is attained, (HA) - 0.112-(H^+). Solve for Ka.
Tuesday, June 15, 2010 at 12:28pm by DrBob222

The titration helps determine the molar concentration of HA: Moles of HA(aq) = moles of OH- used moles of OH- = (0.1500Mmol./L)(0.02500L) = 4.050x10^-3 = mol. HA 4.050x10^-3 mol. HA / 0.02500L = 0.162 M HA Next we find the H+ ion concentration [H+] = 10^-pH = 10^-2.48 = 3....
Monday, October 27, 2008 at 9:38pm by GK

HA + KOH --> H2O + KA mols KOH = M x L = ? mols HA = moles KOH (from the coefficients in the balanced equation. M HA = moles HA/L HA
Monday, March 5, 2012 at 10:52am by DrBob222

mols NaOH = M x L = ? mols HA = mols NaOH (since it's a monoprotic acid) mols HA = grams HA/molar mass HA You know mols HA and grams HA, solve for molar mass HA.
Sunday, April 29, 2012 at 9:42pm by DrBob222

AP Chem
I assume you meant 38.12 mL and not g. mols NaOH = M x L = ? mols monoprotic HA = same. M HA = mols HA/L HA mol HA = grams/molar mass. You have mols and grams, solve for molar mass.
Monday, April 23, 2012 at 5:22pm by DrBob222

The pK of acetic acid is 4.76. Which statement is true at pH 5.0? [A^-] = [HA] [A^-] < [HA] [A^-] > [HA]
Sunday, August 28, 2011 at 4:59pm by Zack

Nobody solve this. I am a cheat!!! Ha ha ha. Jiksha, a website without mods, simply sucks!!!
Monday, June 3, 2013 at 9:32am by keshav

This is just the derivation of the Henderson-Hasselbalch equation. I'll get you started. HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) Now do what b says. Take the log of both sides. log Ka = log(H^+)(A^-)/(HA) or log Ka = log[(A^-)/(HA)] + log(H^+) Multiply through by -1 -log Ka...
Monday, February 28, 2011 at 6:00pm by DrBob222

H2A ==> H^+ + HA^- HA^- ==> H^+ + A^-2 k1 = (H^+)(HA^-)/(H2A) (H^+) = 0.25 from the problem. (HA^-) = 0.25 from the problem. (H2A) = 0.95-.25 = ?? Substitute and solve for k1. k2 = (H^+)(A^=)/(HA^-). (H^+) = 0.25 from the problem. (HA^-) = 0.25 from the problem. (A^=) = ...
Monday, November 15, 2010 at 7:41pm by DrBob222

Ha ha ha Steve!! That's about as far as I got too! It was supposed to end up $400. No idea how though.
Wednesday, February 27, 2013 at 10:27pm by Liz

6th grade_Science/Math
ha ha ha funny josh
Tuesday, January 27, 2009 at 9:45pm by Aaron

%ionization =[(H^+)/(acid)]*100 = ?? We have the (H^+), or we can get it, from pH. I find 3.98 x 10^-8 but you should confirm that. Now, what is (acid)? acid we will call HA. HA ==> H^+ + A^- K = (H^+)(A^-)/(HA) K = (H^+)(A^-)/[HA-(H^+)] Plug in 3.98E-7 for K(from pKa = 6.4...
Sunday, February 13, 2011 at 6:42pm by DrBob222

HA ==> H^+ + A^- Ka = (H^+)A^-)/(HA) Set up an ICE chart and solve for (HA). Then %ion = [H^+)/(HA)]*100 = ? degree ion = %ion/100.
Thursday, October 13, 2011 at 7:44pm by DrBob222

NaOH + HA ==> NaA + H2O mols NaOH = M x L = ? mols HA = mols NaOH (look at the coefficients) M HA = mols HA/L HA
Monday, March 25, 2013 at 1:34pm by DrBob222

Ascorbic acid = HA ................HA ==> H^+ + A^- initial.........0.1.....0......0 change..........-x......x......x equil..........0.1-x.....x.....x Ka = (H^+)(A^-)/(HA) Substitute from the ICE chart and solve for x, then convert to pH.
Wednesday, April 25, 2012 at 7:45pm by DrBob222

ap chem
A 9.00 M solution of a weak acid, HA, has a pH of 1.30 1.) calculate [A-] at equilibrium 2.) calculate [HA] at equilibrium Use pH to get (H^+). Ka = (H^+)(A^-)/(HA) (A^-) = (H^+). (HA) = 9 Plug and chug. Post your work if you get stuck.
Thursday, March 22, 2007 at 4:49pm by Jaron

........HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) This tells you that the 1.070 HA ionized 4.81% which means (H^) = 1.070 x 0.0481 = ? (A^-) = same as (H^+). (HA) = [1.070 - (H^+)] Substitute these values into the Ka expression and solve for Ka.
Wednesday, November 21, 2012 at 12:25am by DrBob222

chemistry - titration
Show that for the titration of a weak acid, HA, with NaOH solution, at half-titer, pH = pKa. Please explain your answer. Thank you. HA + NaOH ==> NaA + HOH Ka for HA = (H^+)(A^-)/(HA) and solve for H^+. (H^+) = Ka*[(HA)/(A^-) When HA is half neutralized, an equal amount of ...
Sunday, August 5, 2007 at 2:11pm by phil

HA = H^+ + A^- Ka = (H^+)(A^-)/(HA) initial: HA = 0.01 H^+ = 0 A^- = 0 change: H^+ = x A^- = x HA = 0.01 - x equilibrium: x = 0.01 x 0.03 = ?? Substitute and solve for Ka.
Monday, May 10, 2010 at 9:15am by DrBob222

HA + NaOH ==> NaA + H2O moles NaOH added = M x L = ?? moles HA must be the same (see the coefficients in the balanced equation). moles HA = grams/molar mass You know moles HA and grams HA, solve for molar mass.
Monday, April 11, 2011 at 5:47pm by DrBob222

f(a+h)= (a+h)^3= (a^2+ah+h^2)(a+h)=a^3+ha^2+h^2a+ha^2+ah^2+h^3 f(a)= a^3 so f(a+h)-f(a)= a^3+ha^2+h^2a+ha^2+ah^2+h^3-a^3 = ha^2+h^2a+ha^2+ah^2+h^3 dividing that by h gives = a^2 + ah+ a^2 +ah^2 + h^2 = 2a^2 as the h approaches zero
Thursday, February 24, 2011 at 11:38am by bobpursley

It tells you it wants you to calculate the Ka for ASA. If we call it HA, then HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) You know (H^+) and you have that right. (A^-) is the same as (H^+). I have used H^+ for H3O^+. For HA, you want to plug in the molarity of the aspirin which ...
Sunday, March 28, 2010 at 8:03pm by DrBob222

College Chemistry
HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) A) pure HA. Set up ICE chart for pure HA, substitute into Ka expression, solve for (H^+) and convert to pH. B) moles HA initially = M x L = ?? moles NaOH at 19.5 mL = M x L = ?? Subtract mols HA - moles NaOH. The difference is the ...
Tuesday, April 13, 2010 at 1:11am by DrBob222

Gen. Chemistry
Call the weak acid HA. HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) pH = 5.20 = -log(H^+) Solve for (H^+), substitute into Ka expression above for H^+ as well as for A^-. For (HA) substitute X-(H^+) and solve for X. I think you can neglect the (H^+) in the X-(H^+) and simply call ...
Sunday, December 5, 2010 at 1:25pm by DrBob222

For salts, like this one, you hydrolyze them; i.e., react them with water. Suppose we have 0.1 M NaA (the salt of NaOH and a weak acid). The general equation is A^- + HOH ==> HA + OH^- Kb = Kw/Ka = (HA)(A^-)/(HA) You know Kw and Ka, (A^-) = 0.1 M and (HA)= (OH^-) = x You ...
Monday, April 14, 2008 at 10:18pm by DrBob222

Let's simplify aspirin formula to HA. HA + H2O ==> H3O^+ + A^- Ka = (H3O^+)(A^-)/(HA) M = (2*0.325)/molar mass HA Set up an ICE chart, substitute into the Ka expression and solve for x. Convert that to pH. I assume you know Ka or you can look it up.
Sunday, March 14, 2010 at 4:39am by DrBob222

chem 2
A buffer solution is made using a weak acid, HA. If the pH of the buffer is 1.0 101 and the ratio of A to HA is 10, what is the pKa of HA?
Thursday, July 18, 2013 at 9:04pm by Brianna

chem 2
A buffer solution is made using a weak acid, HA. If the pH of the buffer is 1.0 101 and the ratio of A to HA is 10, what is the pKa of HA?
Friday, July 19, 2013 at 11:57am by Brianna

If we call aspirin, HA, then the acid dissociates in water as HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) (H^+) = x (A^-) = x (HA) = moles aspirin/L soln. moles aspirin = 2*500/molar mass aspirin. L soln = 0.325 L Solve for x and convert to pH. Post your work if you get stuck.
Wednesday, December 8, 2010 at 11:00am by DrBob222

quadrilateral and angle sums
Kitna practice karoge haa ha ha ha ha ghar jao so jao nahi to gabbar aa jayega haa haha haha
Sunday, February 20, 2011 at 5:40pm by gabbar sing

Feature-Story Ideas
Ya, thats why I shrunk my ideas, leaving campus can be expensive.... Apply for a research grant? HA HA HA (Note Sarcasm) ;D
Wednesday, February 20, 2013 at 7:43am by Ken

HA = acid A^- = conjugate base HA + OH^- ==> H2O + A^- A^- + H^+ ==> HA You can do the other one.
Friday, March 29, 2013 at 2:45pm by DrBob222

Ethnic Diversity
HA ha ha, I am also an Axia student struggling to find the same answer. I think the instructors need to find some new material!
Tuesday, September 2, 2008 at 10:17pm by Amanda

7. At pH 7.4 a weak organic acid with a pKa of 6.4 would be 60% ionized % Ionization =[(H^+)/(acid)]*100 = 3.98 x 10^-8 HA ==> H^+ + A^ K = (H^+)(A^-)/(HA) K = (H^+)(A^-)/[HA-(H^+)] Plug in 3.98E-7
Sunday, February 13, 2011 at 10:54pm by ryan

I assume the unknown acid is a monoprotic one. ..HA...+...NaOH ==> NaA....+ ... H2O 5.55 g..0.005*6.00..0.03 moles..0.03 mols. 0.03 moles/0.750 L = 0.04 M Ka = 1.3E-4 = (H^+)(A^-)/(HA) 1.3E-4 = 5.62E-5)(0.04)/(X) Solve for X = molarity HA. Then M x L = M x 0.750 = moles HA ...
Thursday, November 25, 2010 at 1:55pm by DrBob222

CHEMISTRY FOR DR. BOB or anyone else
A hypothetical weak acid HA, was combined with NaOH in the following proportions: 0.20 mol HA, 0.08 mol NaOH. The mixture was then diluted to a total volume of 1L, and the pH measured. (a) If pH=4.80, what is the pKa of the acid (b) how many additional moles of NaOH should be ...
Friday, May 11, 2007 at 12:38pm by Delli

Crictal Thinking
ha ha ha
Monday, May 18, 2009 at 6:35pm by Anonymous

Check this site.
Sunday, July 20, 2008 at 9:02pm by Ms. Sue

HA ==>H^+ + A^- Ka = (H^+)(A^-)/((HA) You find (H^+) from the pH. The (A^-) = (H^+) and (HA) = 0.1-(H^+) Solve for Ka.
Sunday, February 7, 2010 at 8:46pm by DrBob222

HA ==> H^+ +A^- Ka = (H^+)(A^-)/(HA) You know H^+ (from pH), A^- is the same, and you know HA. Solve for Ka.
Thursday, May 13, 2010 at 5:47am by DrBob222

Proofs and numbers
Wednesday, November 28, 2012 at 10:44am by wISE LECTURER

Use the ΔH°f information provided to calculate ΔH°rxn for the following: ΔH°f (kJ/mol) SO2Cl2 (g) + 2 H2O(l) → 2 HCl(g) + H2SO4(l) ΔH°rxn = ? SO2Cl2(g) -364 H2O(l) -286 HCl(g) -92 H2SO4(l) -814
Wednesday, April 27, 2011 at 8:40pm by yr

a) HA + NaOH ==> H2O + NaA b) moles NaOH = M x L = ?? Using the coefficients in the balanced equation, convert moles NaOH to moles HA. Then M HA = moles HA/L HA. You have moles and L, solve for HA. c) Ka = (H^+)(A^-)/(HA) millimoles HA = mL x M = 25.00*0.3662 = 9.154 mmoles...
Saturday, April 9, 2011 at 2:54pm by DrBob222

We want to prepare 150 mL of 1.0 M HA solution by diluting the stock 8.7 M HA solution with water. What volume of 8.7 M HA solution is needed?
Monday, April 12, 2010 at 11:02pm by billy

HA ==> H^+ + A^- K = (H^+)(A^-)/(HA) = Ka If you know NaOH molarity and NaOH volume, can you calculate the molarity of HA initially? Then knowing (HA), use pH initially to determine (H^+). Of course, (A^-) will be the same (you know this from the ionization equation above. ...
Monday, October 27, 2008 at 9:38pm by DrBob222

Chemistry- Please Help!!
Let's call the unknown acid HA. pH = -log(H^+) -4.5 = log(H^+) (H^+) = about 3E-5 but you need to do it more accurately. ...........HA ==> H^+ + A^- I.........0.1.....0......0 C..........-x.....x......x E........0.1-x....x......x So x = about 3E-5 Substitute into Ka ...
Wednesday, July 31, 2013 at 1:14pm by DrBob222

I can't read the small number in my text but it appears that phenol (C6H5OH) is the smallest which would make Kb the largest. What it means is, we write the equilibrium for the reaction with water as A^- + HOH ==> HA + OH^- Kb = Kw/Ka = (HA)(OH^-)/(HA) So a large Kb means ...
Tuesday, May 4, 2010 at 10:04pm by DrBob222

For H2A it follows the usual form of ionizsation as if you had a monoprotic acid. ..........H2A ==> H^+ + HA^- I.......0.135M....0......0 C.........-x......x......x E........0.135-x..x.......x Ka1 = (H^+)(HA^-)/(H2A) Substitute and solve for x = (H^+) = (HA^-) and convert ...
Saturday, July 21, 2012 at 12:04am by DrBob222

HA + NaOH ==> NaA + H2O is the titration. So at the equivalence point we have NaA which is the conjugate base of HA and you can see that the reverse reaction (which is the hydrolysis) of NaA gives the HA and OH^-. So d.
Thursday, May 26, 2011 at 8:56pm by DrBob222

8th grade
In "Beef Flat"... in comparison to "B Flat" Ha Ha Ha.
Wednesday, April 1, 2009 at 12:35am by Casey

An unknown monoprotic weak acid, HA, has a molar mass of 65.0. A solution contains 2.20 g of HA dissolved 750. mL of solution. The solution has a pH of 2.200. What is the value of Ka for HA?
Tuesday, April 27, 2010 at 7:08pm by Sue

P. Chemistry
HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) Given the pH, convert to (H^+). That is the same as (A^-). You are given either Ka or pKa. That leaves only one unknown, (HA).
Tuesday, February 23, 2010 at 5:37pm by DrBob222

HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) You know HA is 0.490. If it is ionized 2.80%, then H^+ and A^- must be 0.490 x 0.0280 = ??. Substitute into Ka expression and solve.
Monday, April 5, 2010 at 3:03pm by DrBob222

HA ==>H^+ + A^- Ka = (H^+)(A^-)/(HA) Convert pH of 3.00 to (H^+). Plug in that value for (H^+) and (A^-) and for (HA) plug in 0.079-(H^+). solve for Ka, then for pKa. Post your work if you get stuck.
Monday, February 16, 2009 at 4:12pm by DrBob222

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

general, organic and biochemistry
HA is the acid. HA + NaOH ==> NaA + H2O moles NaOH = M x L = 0.6000 x 0.0387 = 0.02322. Therefore, moles HA = 0.02322 M of HA = moles/L. moles = grams/molecular weight m.w. = grams/moles = 4/0.02322 = ?? Then round to 3 significant figures (based on the 38.7)
Tuesday, November 23, 2010 at 6:37am by DrBob222

pH = pKa ONLY when (A^-)/(HA) = 1 Ka = (H^+)(A^-)/(HA). Rearrange to (H^+) = Ka*(HA)/(A^-). When (HA) = (A^-) then (HA)/(A^-) = 1 and (H^+) = Ka*1 or Ka. Take the negative log of both sides. -log(H^+) = -logKa -log(H^+) = pH and -logKa = pKa pH = pKa BUT that is true only when...
Wednesday, October 3, 2012 at 3:34pm by DrBob222

chem-please help!!!!!
If you have a weak acid, HA, and titrate with NaOH, the equation is HA + NaOH ==> NaA + H2O. There is only NaA and H2O at the equivalence point. The A^- of NaA is hydrolyzed (it is a base) as follows: A^- + HOH ==> HA + OH^- Kb = (Kw/Ka) = (HA)(OH^-)/(A^-). You know Kw, ...
Monday, October 18, 2010 at 1:12pm by DrBob222

Convert pH 2.68 to (H^+) with pH = -log(H^+). Then HA --> H^+ + A^- Ka = (H^+)(A^-)/(HA) Substitute for (H^+) and (A^-). For (HA) substitute 0.0169-(H^+)
Saturday, April 21, 2012 at 6:50pm by DrBob222

...........H2A ==> H^+ + HA^- I.........0.0800....0.....0 C.........-x........x......x E........0.0800-x...x.....x k1 = (H^+)(HA^-)/(H2A) Substitute and solve for x = (H^+) and convert to pH. Note that the k2 is about 100 times smaller than k1 so ignoring k2 doesn't cause ...
Friday, October 12, 2012 at 5:42pm by DrBob222

In a field having an area of 3.0 ha is a colony of ground squirrels with 90 members. What is the density of the ground squirrel population? A. 90 ground squirrels. B. 270 ground squirrels/ha. C. 30 ground squirrels/ha. D. 0.030 ground squirrels/ha.
Friday, December 4, 2009 at 9:52pm by Kaled

A^- + HOH ==> HA + OH^- Kb = (HA)(OH^-)/(A^-) You have pH which gives you OH and HA and you have A^-. Calculate Kb
Sunday, June 27, 2010 at 3:15pm by DrBob222

You want to make 1000 ml of an 0.1 M HA buffer so that the pH=pKa of HA. To make this buffer you need to add ____ HA and ____ mL of 1M NaOH and then add enough water to give 1000 mL. HA is a weak acid.
Monday, January 12, 2009 at 5:33pm by Jessica

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

1 ha = 100 X 100 m 1 ha = 0.1 km X 0.1 km 1 ha = 0.01 km ^ 2 1 km ^ 2 = 1 / 0.01 1 km ^ 2 = 100 ha Area of land = 1.2 * 3.6 = 4.32 km ^ 2 4.32 * 100 = 432 ha 432 * 1,500 = 648,000 $
Tuesday, April 30, 2013 at 9:39am by Bosnian

Chemistry (College) URGENT
HA + NaOH ==> NaA + H2O mols NaOH = M x L = ? mols HA = mols NaOH (look at the coefficient in the balanced equation). mols HA = grams HA/molar mass HA. You know mols and grams, solve for molar mass. If the pH at the half way point is 3.86, then the pKa is 3.86. pKa = -log ...
Monday, October 29, 2012 at 6:45pm by DrBob222

ha ha ha!! :)
Thursday, August 21, 2008 at 6:19pm by Cindy

ha is an area measure km is a length measure 1 ha = 10,000m^2 850 ha = 8,500,000 m^2 1km = 1000m 1km^2 = (1000m)^2 = 1,000,000 m^2 so, 850 ha = 8.5 km^2 so, your answer is correct, but your conversion factor should have been written 1km^2 / 100ha
Wednesday, October 17, 2012 at 10:43pm by Steve

HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) Convert pH to H^+, substitute that for H^+ and A^- in the equation for Ka. For HA, substitute 0.23-H^+ and solve for Ka.
Tuesday, November 16, 2010 at 8:52pm by DrBob222

HA ==> H^+ + A^- (H^+)(A^-)/(HA) = Ka moles HA = moles/molar mass = 2.20/65 = 0.0338 (HA) = (0.0338/0.750) = 0.0451 M pH = 2.200 2.200 = -log(H^+) (H^+) = 0.00631 Substitute into Ka expression. (0.00631)(0.00631)/(0.0451-0.00631) = Ka. You can finish.
Tuesday, April 27, 2010 at 7:08pm by DrBob222

Chem 1A
Look at HA ionization. HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) Solve for (H^+) = Ka(HA)/(A^-) Note that 700 mL is exactly half way to the equivalence point; therefore, we know (HA) untitrated = (A^-) from the half titrated. Thus pKa = pH. At th stoichiometric point you know ...
Thursday, March 15, 2012 at 10:04pm by DrBob222

With Ka1 and Ka2 as close as they are, there is some interaction between the two but I will assume you can take the easy way out, assume there is no interaction, and solve the problem this way. The REAL answer won't be that far away (perhaps 3-5%). ..........H2A ==> H^+ + ...
Sunday, October 20, 2013 at 5:14pm by DrBob222

Use the Henderson Hasselbalch equation twice. First set of conditions: pH=9.0 pka=9.6 Solve for the ratio pH=pka+log[A-/HA] 9.0=9.6+log[A-/HA] 10^(9.0-9.6)=[A-/HA] 0.25=[A-/HA] Meaning 25% of the solution is protanated, or 0.025 moles is A- and 0.075 moles HA Second set of ...
Wednesday, February 27, 2013 at 9:48pm by Devron

pH = 2.73 = -log(H^+). Solve for (H^+) and I get something like 1.86E-3 but you need to confirm that. Let HA = fatty acid. .........HA --> H^+ + A^- .........x.......0....0 equil....x......1.86E-3..1.86E-3 Ka = (H^+)(A^-)/(HA) Substitute into Ka expression and solve for x.
Saturday, December 3, 2011 at 3:41am by DrBob222

AP Chemistry
I have tried several times to post this but the site has been having some problems. Maybe this time. We take a salt NaA and throw it into some water. The Na^+ doesn't hydrolyze but the A^- does. I will omit all of the charges to make this simpler to type. A + HOH ==> HA + ...
Wednesday, February 17, 2010 at 8:11pm by DrBob222

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

Let HA = benzoic acid ----------- moles NaOH(initial) = (0.015L)(0.50 mol/L) = 0.0075 mole moles HA(initial) = (0.0300L)(0.50 mol/L) = 0.015 mol OH^- + HA --> H2O + A- moles of A- formed = 0.0075 (same as initial OH-) moles HA remaining = 0.0150- 0.0075 =0.0075 mol [A-] = ...
Tuesday, July 28, 2009 at 12:32am by GK

pH = -log(H^+) -3.25 = log(H^+) (H^+) = 5.62E-4 ............HA ==> H^+ + A^- initial.....x.......0.....0 equil.......x...5.62E-4..5.62E-4 Ka = (H^+)(A^-)/(HA) Substitute (H^+), (A^-) and Ka, and solve for (HA) = x
Thursday, December 1, 2011 at 12:45pm by DrBob222

AP Chemistry
HA ==> H^+ + A^- Solve this equation to determine the (HA) at pH 4.50. As a separate problem, plug in pH 5.000 and determine the (HA) for that value. Then use the dilution formula using 67.0 mL of the first to determine how much to dilute it. c1v1 = c2v2 Post your work if ...
Tuesday, February 7, 2012 at 7:19pm by DrBob222

se the information provided to determine ΔH°rxn for the following reaction: ΔH°f (kJ/mol) 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH°rxn = ? Fe2O3(s) -824 Fe3O4(s) -1118 CO(g) -111 CO2(g) -394
Wednesday, April 27, 2011 at 8:40pm by yr

For 1, pH = -log(H^+). Plug and chug. If you don't understand how to do this, explain your exact problem and I can help you through it. (It may be so simple that you just don't know how to handle the calculator.). #2. No arrow. I don't know the reactants from the products. But...
Saturday, June 12, 2010 at 5:01pm by DrBob222

Call the weak acid HA. (HA) = mols/L = 0.187/0.725 = about 0.26M but you should be more accurate than that. .......HA --> H^+ + A^- I.....0.26M....0.....0 C.....-x.......x.....x E....-0.26-x...x.....x Ka = 1.7E-3 = (H^+)(A^-)/(HA) Substitute into Ka expression and solve for...
Tuesday, April 16, 2013 at 11:54am by DrBob222

chemistry plse help
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction: P4(g) + 10 Cl2(g) → 4PCl5(s), ΔH°rxn = ? Given: PCl5(s) → PCl3(g) + Cl2(g), ΔH°rxn= +157 kJ P4(g) + 6 Cl2(g) → 4 PCl3(g), ΔH°rxn = -1207 kJ
Wednesday, April 27, 2011 at 8:05pm by yr

American Government
Friday, December 25, 2009 at 11:58pm by Ha ha Engineer

Help, Math 8th grade!!!!!
sorry lol, for the name ha ha
Tuesday, November 13, 2012 at 2:28pm by Esther

Close but not quite there. If the equivalence point is at 4.80 mL, then go to 2.40 mL (the half-way point) and the pH at that point equals the pKa (not Ka). The theory is simple enough. A weak acid HA ==> H^+ + A^- and Ka = (H^+)(A^-)/(HA). Solve for (H^+)=Ka*(HA)/(A^-). At...
Monday, March 21, 2011 at 12:10am by DrBob222

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