Friday
April 18, 2014

Search: Ka for benzoic acid, C6H5COOH, 6.5*10^-5. CALCULATE THE pH OF SOLUTION AFTER ADDITION OF 10.0, 20.0 ML OF 0.10M NaOH TO 40.0ML OF 0.10 M BENZOIC ACID

Number of results: 99,219

Prcts chemistry
Calculate the pH of 0.0115M acetic acid solution at 25 degrees celcius with Ka = 1.75*10^-5
Thursday, September 22, 2011 at 5:12pm by Gift

ap chem
The pH of a 0.100 M solution of a weak acid HA is 4.94 . Calculate Ka for the acid.
Monday, February 14, 2011 at 1:29am by Anonymous

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
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

Chem 2
Using the reaction HB<=> H+ + B- and the Ka for that expression, calculate the Ka given the initial concentration of weak acid to be 0.150 M and the pH of the solution to be 2.79.
Friday, July 27, 2012 at 3:20pm by Myssey

chemistry
What you may mean is to ask if crude benzoic acid has a higher or lower melting point than recrystallized benzoic acid. The purer benzoic acid has a higher melting point than impure benzoic acid.
Thursday, July 21, 2011 at 12:58am by DrBob222

Chemistry
Buffers- Common Ion effect IS this Correct Calculate the pH of a aqueous solution containing 0.15 M HNO2 and 0.20 M NaNO2 (aq). The Ka of nitrous acid is 4.0*10^-4. HNO2 + H20 <--> NaNO2- + H30^+ Inital 0.15 0.20 X Final 0.15 0.20 Ka= [H30+][NaNO2]/ [HNO2] 4.0*10^-4 = [...
Tuesday, August 11, 2009 at 8:17am by Saira

Chemistry
Hydrogen cyanide is a weak acid with Ka = 4.9 x 10-10. Calculate the pH of a solution (to 2 decimal places) containing 2.28 g of Sr(CN)2 per 158 mL.
Monday, January 7, 2008 at 8:34pm by Andrew

Chemistry
Using 6.14 x 10^-5 for Ka for benzoic acid, I obtained 47.8% ionization.
Saturday, February 14, 2009 at 6:01pm by DrBob222

Chemistry
HOBr (aq) <----> H+ (aq) + OBr- (aq), Ka = 2.3 x 10^-9 Hypobromous acid, HOBr, is a weak acid that dissociates in water, as represented by the equation. (a) Calculate the value of [H+] in a solution of HOBr that has a pH of 4.95. Ans: 1.1 * 10^-5 M (b) Write the ...
Tuesday, April 10, 2012 at 7:22pm by Confused

CHEM!
Okay well I suddenly got confused for a lab I did We took pH meter than we calibrate it. Then we measure the pH of the acetic acid solution, which was used to determine the Ka of the acetic acid using the first method. We made a 1M solution of NaOH which was used to titrate 50...
Wednesday, May 9, 2007 at 6:05pm by Linda

Chemistry
Calculate the pH of a 0.10 M NaHX salt solution. Ka (H2X) = 2.1 x 10-2 Ka (HX-) = 4.5 x 10-8 Can someone explain the steps to do this please?
Friday, October 21, 2011 at 8:05pm by ellen

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 (A level)
I was given a Chemistry question on the titration of a strong base being added to a weak acid. The question was: "Calculate the pH in a titration when 10.0cm^3 of a 0.10moldm^-3 solution of NaOH is added to a 10.0cm^3 of 0.25moldm^-3 solution of ethanioc acid? (Ka = 1.76 x 10...
Saturday, December 6, 2008 at 9:49am by Benjy

Chemistry(Please check)
an acid with the equilibrium concentration is listed below: 1) HA + H2O = H3O^+ + A- HA= 10^-1 M, H3O=10^-3M, A=10^-3M Calculate the Ka and pKa for the acid. Ka=[10^-3][10^-3] / [10^-1] = 1.00 X 10^-5 pKa = -log[1.00 X 10^-5] = -1.00 Did I do this correctly for the Ka and pKa?
Thursday, September 20, 2012 at 6:36pm by Hannah

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
Monochloroacetic acid, HC2H2ClO2, is a skin irritant that is used in "chemical peels" intended to remove the top layer of dead skin from the face and ultimately improve the complexion. The value of Ka for monochloroacetic acid is 1.35 10-3. Calculate the pH of a 0.22 M ...
Wednesday, November 23, 2011 at 7:08pm by Kaleen

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

Chmeistry
A lactic acid/lactate ion buffer solution contains 0.19 M HC3H5O3 and 0.78 M C3H5O3-, respectively. The Ka value of lactic acid is 1.4*10^-4. Calculate the pH of this buffer.
Saturday, June 5, 2010 at 12:59pm by Lynne

Chemistry
A lactic acid/lactate ion buffer solution contains 0.19 M HC3H5O3 and 0.78 M C3H5O3-, respectively. The Ka value of lactic acid is 1.4*10^-4. Calculate the pH of this buffer.
Saturday, June 5, 2010 at 1:00pm by Lynne

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
Benzoic acid is a solid. Nitrobenzene is a liquid; therefore, mols benzoic acid is 12.2 g/molar mass benzoic acid. The solvent is nitrobenzene and you have 0.250 kg.
Monday, October 27, 2008 at 12:02am by DrBob222

Chemistry
In a experiment to determine the molecular weight and the Ka for ascorbic acid (vit. c.) a student dissolved 1.3713g of the monoprotic acid in water to make 50 mL of solution. The pH was monitored throughout the titration. The equivalence point was reached when 35.23 mL of the...
Saturday, May 19, 2007 at 6:57pm by Tri

Chemistry
A very old and tired , grey haired AP Chem instructor wanted to determine the Ka of an unlabelled monoprotic acid in his stockroom. He dissolved an unknown amount of acid in an unknown amount of water and proceeded to titrate the sample with a solution of NaOH of unknown ...
Sunday, February 24, 2013 at 8:42pm by Kat

Chemistry
A very old and tired , grey haired AP Chem instructor wanted to determine the Ka of an unlabelled monoprotic acid in his stockroom. He dissolved an unknown amount of acid in an unknown amount of water and proceeded to titrate the sample with a solution of NaOH of unknown ...
Sunday, February 24, 2013 at 9:19pm by Summer

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
The Ka for hydroflouric acid is 4.5 *10^-4 at 298K. Calculate the amount of sold NaF that is required toprepare a 250. mL acidic buffer with pH = 2.75. The inital molarity of hydroflouric acid is .250M but you have only 75mL.
Thursday, April 7, 2011 at 12:04am by Jake

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
I said I would find your error but I've changed my mind after I see how you worked the problem. Have you seen the Henderson-Hasselbalch equation? Have you used it? Why not use it here? It is MUCH simpler and it uses exactly the same chemistry; just a different form of it. pH...
Friday, March 18, 2011 at 8:57pm by DrBob222

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

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

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
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

Chemistry III
Assume you dissolve 0.240 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 1.16 102 mL of solution and then titrate the solution with 0.153 M NaOH. C6H5CO2H(aq) + OH -(aq) C6H5CO2-(aq) + H2O(l) (a) What was the pH of the original benzoic acid solution? (b) ...
Tuesday, March 27, 2012 at 11:09pm by Kelly

Chemistry
1. HCl (strong acid completely split into ions) *Find the concentration after dilution using: V1M1 = V2M2 (8.00mL)(0.10 M) = (100.0 mL)(M2) *M2 = (8.00mL)(0.10 M) / (100 mL) = ?______M *pH = -log(M2) 2. HC2H3O2 (weak acid only partially dissociated into ions): *Find the ...
Wednesday, August 20, 2008 at 3:26am by GK

chemistry
URGENT!!! How many grams of sodium benzoate (NaC6H5CO2) must be added to 500 mL of 0.15 M benzoic acid (HC6H5CO2) to make a buffer solution with pH of 5.0? Assume no volume change upon sodium benzoate dissolution. Ka = 6.46 x 10-5 How would I go about this question?
Wednesday, March 26, 2014 at 1:29pm by Rainbow

AP Chemistry
A very old and tired , grey haired AP Chem instructor wanted to determine the Ka of an unlabelled monoprotic acid in his stockroom. He dissolved an unknown amount of acid in an unknown amount of water and proceeded to titrate the sample with a solution of NaOH of unknown ...
Sunday, February 24, 2013 at 2:42pm by Sarah

chem
The procedure for finding pH at the beginning is correct but you made a math error somewhere. It appears you may have used Kb and not Ka. 1.5 x 10^-5 = (H^+)^2/0.105 and solve for (H^+). The pH is between 2 and 3. Volume to reach the equivalence point is mL x M = mL x M for ...
Monday, March 1, 2010 at 11:38pm 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 (H...
Thursday, July 19, 2007 at 1:47am by Jess

chemistry
A student performed the nitration of Benzoic Acid. She/he started with 7.2 g of starting material and obtained 4.8 g of product. Assuming Benzoic Acid is the limiting reagent; calculate the theoretical and percent yields. Show the equation and all steps of your calculations.
Wednesday, January 18, 2012 at 12:25pm by marisol

Chemistry 122
A solution of nicotinic acid, HC6H4NO2, of concentration 0.012 M has a pH of 3.39 at 25 degrees Celsius. Calculate Ka for this acid.
Monday, November 22, 2010 at 10:19am by Marlynda

Chemistry
This problem relies on the Henderson-Hasselbalch Equation which relates the pKa of a buffer to changes in the solution such as the addition of strong acids or strong bases. So the first thing to do in this problem is to find the pKa of benzoic acid. That can be done using a ...
Thursday, November 1, 2012 at 1:00pm by Anonymous

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
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 210−11 1. 7 2. 10 3. 13 4. 5 5. 1
Thursday, February 25, 2010 at 10:44pm by hello

Chm 2
The way you do these is to convert those Ka values to pKa = -log Ka. Then the best acid/conjugate base is the acid with a pKa = +/- 1 of the pH you want for the buffer. For example, pKa for acetic acid is 4.74 and that would be the best acid (conjugate base is acetate ion) to ...
Tuesday, October 15, 2013 at 12:29am by DrBob222

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

Chemistry...Attn DrBob222
Thanks for your earlier thoughts. Am still working on the problem and would like to know if you have any thoughts on my approach below: If I have a set of data that gives the apparent partition coefficient as a function of pH, over a range of pH 2-10, how do I calculate the Ka...
Friday, March 12, 2010 at 8:16pm by Renee

Chemistry
C6H5COOH = HB ........HB ==> H^+ + B^- I.......0.048...0....0 C........-x.....x....x E.....0.048-x...x.....x Ka = ? = (H^+)(B^-)/(HB) You should find Ka in your text or notes. Substitute and solve for x = (H^+) and convert that to pH, pOH, and (OH^-).
Monday, December 9, 2013 at 4:46pm by DrBob222

Chemistry
The correct answer to your question is that you must recognize the kind of solution you have. Sometimes that is easier said than done, especially until you get some practice at it. Practice makes perfect. 1. If it is a salt (a,b,d), the Bronsted-Lowry Theory tells you. In the ...
Saturday, March 13, 2010 at 6:03pm by DrBob222

chemistry
At 200 mL, you are at the equivalence point. acetic acid KOH ==> Kacetate + H2O acetic acid initially = 0.02 moles KOH added 0.2 L x 0.1 M = 0.02 moles. So what do we have in solution. It is Kacaetate (potassium acetate) in 300 mL water. What determines the pH of salts in ...
Thursday, July 23, 2009 at 11:12pm by DrBob222

Chemistry!! Please Help!
A 0.285 M solution of the sodium salt, NaA, of the weak monoprotic acid, HA, has a pH of 9.65. Calculate Ka for the acid HA. I think I am having issues with the equation and then the math part. I figured out what 'x' was from the pH, which I believe is 2.2x10^-10, but when I ...
Friday, February 6, 2009 at 3:38pm by Blair

Chemistry
Use the Henderson-Hasselbalch equation to calculate the pH of a buffer prepared by adding 10mL of acetic acid to 20mL of 0.10M sodium acetate (Ka= 1.75 x 10^5 for acetic acid). Please showing working out, as am confused as to how to approach this question. Thanks in advance ...
Sunday, May 9, 2010 at 7:35am by Lois

ap chemistry
What is the concentration of H+ ions in a buffer solution containing 0.305 M benzoic acid (with ionization constant 6.5 10−5)and 0.834 M sodium benzoate (a salt of benzoic acid)?
Tuesday, February 4, 2014 at 6:56pm by gabriella

A few chemistry
I got the rest of the homework, but these are confusing me so much...ugghhhh. 1. The pH of a .400 M solution of iodic acid, HlO3, is .726 at 25 degrees C. What is the Ka(acid constant) at this temperature? 2. The pH of a .150 M solution of HClO is found to be 4.55 at 25 ...
Thursday, March 22, 2007 at 10:12pm by Trish

Chemistry
Find the pH of mixture of acids. 0.185 M in HCHO2 and 0.225 M in HC2H3O2 Im using an ice chart of weak acid and putting in strong acid in H+ initiAL concentration. I've done the problems many different ways but cannot seem to get the right answer help please?Answer is pH of 2....
Sunday, February 26, 2012 at 10:02pm by L.Bianchessi

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
a 5.55g sample of a weak acid with ka=1.3*10^-4 was combined with 5.00ml of 6.00 M NAOH and the resulting solution was diluted to 750mL. The measured pH of the solution was 4.25. what is the molor mass of the weak acid. if used the formula pH=kpa+log(base/acid) Ph=3.89+log(6....
Thursday, November 25, 2010 at 1:52pm by help

chemistry
a 5.55g sample of a weak acid with ka=1.3*10^-4 was combined with 5.00ml of 6.00 M NAOH and the resulting solution was diluted to 750mL. The measured pH of the solution was 4.25. what is the molor mass of the weak acid. if used the formula pH=kpa+log(base/acid) Ph=3.89+log(6....
Thursday, November 25, 2010 at 1:55pm by help

Chemistry
If the Ka of a monoprotic weak acid is 4.3*10^-6, what is the pH of a 0.16 M solution of this acid?
Sunday, July 24, 2011 at 12:35am by Josh

Chemistry
If the Ka of a monoprotic weak acid is 9.0 10-6, what is the pH of a 0.26 M solution of this acid?
Tuesday, October 25, 2011 at 8:39pm by Sevy

science
If the Ka of a monoprotic weak acid is 8.3 10-6, what is the pH of a 0.28 M solution of this acid?
Monday, July 22, 2013 at 7:16pm by candace

chemistry
Let HB = benzoic acid. HB ==> H^+ + B^- Ka = (H^+)(B^-)/(HB) Set up an ICE chart. Convert pH=2.90 to (H^+). Substitute into Ka expression for (H^+) and for (B^-) (they are equal). Solve for (HB) which will be in units of M = moles/L. You know M, you know how many L you want...
Monday, April 11, 2011 at 4:57am by DrBob222

chemistry
the ph of a 1.00*10^-2 M solution of cyanic acid (HOCN) is 2.77 at 25 C. Calculate Ka and pKa for HOCN from this result.
Saturday, June 30, 2012 at 10:02pm by FB1907

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. Ac^- is ...
Monday, November 15, 2010 at 12:20am by DrBob222

Chemistry 2 Lab
Calculate the degree of ionization of acetic acid in solution 1 through 3? All I am giving is the pH... pH of solution 1= 3 pH of solution 2= 3.5 pH of solution 3= 5 Then, I did this... HC2H3O2 (aq) + H2O (l) equals H3O+ (aq) + C2H3O2- (aq) Ka= 1.7 E -5 (according to chart in ...
Tuesday, March 19, 2013 at 6:50pm by Robby

Organic Chemistry
Would you expect benzoic acid to react with NaHCO3. C6H5COOH + NaHCO3 ==> CO2 + H2O + C6H5COONa
Wednesday, November 17, 2010 at 2:15pm by DrBob222

science
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 4:21pm by Jodi

Chemistry
so i had to calculate the pH of a 100ml solution of .10M acetic acid (ka=1.8x10(-5) I came out to 2.87. But then, they want the pH if 50ml HCL is added. So i thought...Ka = x2/.15M, therefore the square root of .15M x 1.8x10(-5) = .00164(concentration). So -log(.00164) = 2.78...
Thursday, October 20, 2011 at 1:21pm by Kevin

Chemistry
A 0.10 M solution of a weak acid has a pH of 3.5 at 25%C. What is the equilibrium constant, Ka, for this acid?
Sunday, October 10, 2010 at 10:03am by Tara

Chemistry
First, a correction. The H-H equation is pH = pKa + log (base/acid); your fraction is invered incorrectly. Here are both ways: HBrO ==> H^+ + BrO^- Ka = 2 x 10^-9 = (H^+)(BrO^-)/(HBrO) Plug in (BrO^-) 6.50grams/molar mass NaBrO and that divided by 0.110 L. (HBrO) = 0.5 M ...
Tuesday, March 16, 2010 at 7:46pm by DrBob222

Organic Chemistry
I am suppose to calculate the theoretical amount (volume) of 3M sodium hydroxide needed to convert 1.5 g of benzoic acid to its salt. Also calculate the volume of 12M HCl needed to convert the sodium benzoate back to benzoic acid. how do I go about doing this? Thanks
Tuesday, May 18, 2010 at 1:25pm by Jake

Chemistry
The pH of a 1 .00 x 10-2 M solution of cyanic acid(HOCN) is 2.77 at 25 degrees. Calculate Ka and pKa for HOCN from this result.
Monday, March 11, 2013 at 8:58pm by Annie

chemistry
You weigh a sample of a monoprotic unknown acid and dissolve it in 50.00 mL of distilled water. Exactly half of this solution is titrated with Sodium Hydroxide to the phenolphthalein end point. The pH of the other half of the original solution is measured with a pH meter. The...
Tuesday, July 30, 2013 at 4:46am by faruk

chemistry
Let's let HPh equal phenol with H representing the hydrogen that ionizes and Ph representing the remainder of phenol. HPh ==> H^+ + Ph^- Write the Ka expression. Ka = 1 x 10^-10 = [(H^+)(Ph^-)]/(HPh) Before the ionization (HPh) = 0.106 M. Before the ionization (H^+)=(Ph^-)=...
Thursday, January 10, 2008 at 7:22pm by DrBob222

Chemistry
The pH of 0.15 mol/L of hydrocyanic acid is 5.07. Calculate Ka of hydrocyanic acid.
Saturday, November 26, 2011 at 5:45pm by Brandon

Chemistry
Enough of a monoprotic acid is dissolved in water to produce a 0.0195 M solution. The pH of the resulting solution is 6.24. Calculate the Ka for the acid. Also, the answer is not 1.7 * 10 ^ -11 The response is: You treated the initial concentration as 0. In this case, the ...
Sunday, April 22, 2012 at 8:50am by chem1

Chemistry
Call benzoic acid HB and sodium benzoate NaB. .............HB ==> H^+ + B^- initial....0.15M....0.....0 change.......-x......x.....x equil.....0.15-x......x....x I used 6.14E-5 for Ka. 6.14E-5 = (H^+)(B^-)/(HB) 6.14E-5 = (x)(x)/(0.15-x) x = (H^+) = about 0.003 which is (0....
Monday, March 12, 2012 at 2:10am by DrBob222

Chemistry
If you are looking for percent recovery benzoic acid as a percent of the original sample it is (0.05/0.2)*100 = 25% but that may not be percent recovery of the benzoic acid since you don't know how much benzoic acid was in the sample initially. If you can assume the fluorenone...
Friday, October 25, 2013 at 4:09am by DrBob222

College Chemistry
You weigh a sample of a monoprotic unknown acid and dissolve it in 50.00 mL of distilled water. Exactly half of this solution is titrated with Sodium Hydroxide to the phenolphthalein end point. The pH of the other half of the original solution is measured with a pH meter. The...
Tuesday, April 9, 2013 at 4:31pm by Kim

pH- really hard one
What is the pH of the solution created by combining 1.00 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)? NaOH + HCl ==> NaCl + H2O So all the NaOH is neutralized leaving NaCl, which will not affect the pH either way, ...
Sunday, April 15, 2007 at 7:39pm by Christine

chemistry
Given 1.00L of a buffer containing 0.338 M CH3COOH and 0.093 M CH3COONa. Ka(acetic acid) = 1.8 x 10-5. Calculate the pH after the addition of 0.005 moles NaOH.
Sunday, March 18, 2012 at 7:39pm by cathy

chemistry
Given 2.00L of a buffer containing 0.338 M CH3COOH and 0.093 M CH3COONa. Ka(acetic acid) = 1.8 x 10-5. Calculate the pH after the addition of 0.004 moles NaOH.
Wednesday, March 21, 2012 at 11:23pm by reichel

school
What is the pH of 0.1 M formic acid solution? Ka=1.7e10-4? What is the pH value of buffer prepared by adding 60 ml of 0.1 M CH3COOH to 40 ml of a solution of 0.1 M CH3COONa? What is the pH value of an acetate buffer (pK=4.76) prepared by adding 20 ml of 1 M HCl to 100 ml of a...
Sunday, December 11, 2011 at 7:58am by alshammari

pH- from saturday
i posted this question previously and got help however im still having some confusion. can u clarify ? what's the pH of the acid HC2H3O2 with concentration of .15? Bob Pursley worked that problem for you and gave it as [x^2/0.15] = Ka. What is it you don't understand? You know...
Sunday, April 15, 2007 at 7:48pm by Christine

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

Plz -- Chem Help
Posted by Rushi on Sunday, March 15, 2009 at 3:35pm. 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 ...
Wednesday, March 18, 2009 at 10:54am by A

chemistry
Lactic acid is a weak monoprotic acid. If a 0.10 M solution of it has a pH of 2.44, what is the dissociation constant (Ka) for lactic acid. I got 1.3x10^-4? Is that right?
Monday, October 3, 2011 at 5:46pm by person

Chemistry - pH
a) 0.35M HCl (monoprotic strong acid) produces a 0.35M [H+] solution. pH = -log(0.35) c) 0.35M NaOH (strong base) acid produces a 0.35M [OH-] solution. pOH = -log(0.35) pH = 14-pOH b) 0.35M HC2H3O2 (WEAK ACID) does not dissociate completely. Its Ka = 1.8x10^-5 [H+] = sqrt[(Ka...
Monday, September 14, 2009 at 10:12am by GK

Chemistry
A 0.150 M solution of nitrous acid (HNO2) is made. Ka = 4.5 x 10-4 1. Show the equilibrium which occurs when this acid is dissolved in water. 2. What is the pH of the solution? Show all work clearly. 3. 100.0 mL of the solution is titrated against 0.150 M NaOH. What is the pH ...
Friday, April 6, 2012 at 4:57pm by Tiri

Chemistry
A 0.150 M solution of nitrous acid (HNO2) is made. Ka = 4.5 x 10-4 1. Show the equilibrium which occurs when this acid is dissolved in water. 2. What is the pH of the solution? Show all work clearly. 3. 100.0 mL of the solution is titrated against 0.150 M NaOH. What is the pH ...
Friday, April 6, 2012 at 5:10pm by Tiri

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

Chemistry
Enough of a monoprotic acid is dissolved in water to produce a 0.0169 M solution. The pH of the resulting solution is 2.68. Calculate the Ka for the acid.
Saturday, April 21, 2012 at 6:50pm by Lee

chemistry
enough of a monoprotic acid is dissolved in water to produce a 0.0111M solution. The pH of the resulting solution is 2.62. Calculate the Ka for the acid.
Friday, October 12, 2012 at 3:33pm by airo

Chemistry
Enough of a monoprotic acid is dissolved in water to produce a 0.0190 M solution. The pH of the resulting solution is 2.33. Calculate the Ka for the acid.
Tuesday, March 26, 2013 at 1:15pm by Ann

Chemistry
Enough of a monoprotic acid is dissolved in water to produce a 1.47 M solution. The pH of the resulting solution is 2.64. Calculate the Ka for the acid.
Wednesday, April 10, 2013 at 8:54pm by Julia

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
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
I have no clue, please help and explain! =) Find the pH of a 0.150 M solution of a weak monoprotic acid having Ka=1.210−5. Find the percent dissociation of this solution. Find the pH of a 0.150 M solution of a weak monoprotic acid having Ka=1.410−3. Find the ...
Monday, December 6, 2010 at 7:33pm by Lisa, Help needed!

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