Determine the pOH of a solution after 0.1 moles of NaOH is added to 1L of a solution containing 0.15M CH3COOH ?
Moles acid = 0.15 moles salt = 0.2 CH3COOH + OH- >> CH3COO- + H2O moles acid = 0.15 - 0.1 = 0.05 moles salt = 0.2 + 0.1 = 0.3 pKa = 4.74 pH = 4.74 + log 0.3 / 0.05 =5.5 pOH = 8.5
Calculate the pH after 0.014 mol NaOH is added to 1.00 L of each of the four solutions.?
In (a) you start with 0.1 M x 1 L = 0.1 mol of acid. You add 0.014 mol of base. Your final NaOH concentration is zero, and your final acid concentration is 0.100 - 0.014 = 0.086, and you have formed 0.014 mol of the conjugate base. pH = pKa + ([A(-) / [HA]) = 4.89 + log (0.014 / 0.086) = 4.10 In (b) you have base to start with, but its a weak base. The strong base will dominate the pH. pOH = -log(0.014) = 1.85 pH = 14 - pOH = 14 - 1.85 = 12.1 In (c) it's the same as in (b). In (d) only the acid will react, so you get 0.086 mol of HA and 0.114 mol of the base (some was already there, plus the stuff you form in the reaction of HA with NaOH). pH = pKa + ([A(-) / [HA]) = 4.89 + log (0.114 / 0.086) = 5.01
AP chem help!! Calculate the pH after .02 moles NaOH have been added...?
Calc pH after .02 moles NaOH have been added to .1 M HONH2 (Kb= 1.1 x 10^-8) thanks! I know HONH2+ H2O = HONH3+ + OH- but where to go from there??? Kb= [HONH3+] [OH-] --------------------- [HONH2]
Calculating pH during titration?
100.0 mL of 0.100 M acetic acid is placed in a flask and is titrated with 0.100 M sodium hydroxide. An appropriate indicator is used. Ka for acetic acid is 1.7 s 10^-5. Calculate the pH in the flask at the following points in the titration. a) When no NaOH has been added b) After 25.0 ml of NaOH is added c) After 50.0 ml of NaOH is added d) After 75.0 ml of NaOH is added e) After 100.0 ml of NaOH is added f) After 300.0 ml of NaOH is added I just want to make sure that my math is right... For A- I got a pH of 1 by doing the -log of ((.100 M x .100 L) - (.100 M x 0 L)) // .100 L. I used the same technique to get b) 1.22 c) 1.48 d) 1.85 Is that correct? And then I'm having a hard time with e because with what I'm doing I get -log(0) which of course is undefined.... Help would be appreciated!
What is the pH when 32.5 mL of 0.0170 M NaOH has been added to 25.0 mL of 0.0655 M acetic acid?
Calculate how many moles of acid you have, and how many moles of base. See which is in excess, and by how much, then calculate pH. moles of NaOH = 0.0325 L x 0.017 mol/L = 0.00055 moles moles of acetic acid = 0.025 L x 0.0655 mol/L = 0.00164 moles .00164 - .00055 = 0.001 moles of acetic acid in excess total volume = 32.5 mls + 25 mls = 57.5 mls 0.001 moles/0.0575L = 0.017 M un-neutralized acetic acid HAc ==> H + Ac Ka = 1.8x10^-5 = [H][Ac]/[HAc] = X^2/0.017 X^2 = 3.15x10^-7 and X = [H] = 5.6x10^-4 pH = -log [H] = -log 5.6x10^-4 = 3.25 I don't get 4.45, but rather get 3.25. I simplified the quadratic but I doubt that makes the difference.
What is the pH when 100 ml of 0.1 N NaOH is added to 150 ml of 0.2 M Acetic acid ,?
Molarity of NaOH = normality of NaOH moles NaOH =0.100 L x 0.1 M = 0.010 moles acetic acid = 0.150 L x 0.2 M = 0.030 the reaction is CH3COOH + OH- = CH3COO- + H2O moles acetic acid in excess = 0.030 - 0.010 = 0.020 moles acetate formed = 0.010 total volume = 0.250 L [acetic acid]= 0.020/ 0.250=0.080 M [acetate]= 0.010/ 0.250=0.040 M pH = pKa + log [acetate]/ [acetic acid] ( Handerson-Hasselbalch equation) pH = 4.76 + log 0.040/0.080=4.46
Consider a buffer solution containing 0.1 mole each of acetic acid and sodium acetate in a 0.1 L of solution. 0.01 mole of NaOH is gradually added to this buffer solution. What will be the new [H+] ion concentration in the resulting solution?
Initial concentration of acetic acid as well as sodium acetate is 1 .( molarity = moles / vol in liters)When you add NaOH , being basic it will react with acetic acid forming acetic acid and water.CH3COOH + NaOH = CH3COONa + H2OConc of NaOH added is 0.1 so 0.1M will react with CH3COOH.After reaction the conc of CH3COOH decrease by 0.1 hence it becomes 0.9 ( initial was 1.)And conc of CH3COONa increase by 0.1 becoming 1.1 ( initial was 1)Now this is acid buffer so you can use the formulapH = pKa + log [conc of salt/ conc of acid]New Conc of salt CH3COONa = 1.1New Conc of Acid CH3COOH = 0.9If you know the value of pKa , you can calculate pH hence you can find [H+] by taking anti log.I hope you get it.
A student adds 15.0 mL 2.00 M Acetic Acid and 20.0 mL 1.50 M NaOH. Determine the pH of the resulting solution.?
find moles (0.0150 L) (2.00 mol / Litre) = 0.0300 moles Acetic Acid (0.0200 L) (1.50 mol / Litre) = 0.0300 moles of NaOH. by the equationL: 1 mole H C2H3O2 & 1 mole NaOH --> 1 mole Na C2H3O2 & 1 H2O equal numbers of moles react so 0.0300 moles Acetic Acid neutralize 0.0300 moles of NaOH, producing 0.0300 moles Na C2H3O2 that Na C2H3O2 is now in 35 ml, find molarity of the acetate ion produced (0.0300 moles Na C2H3O2) / (0.0350 Litres) = 0.8571 Molar (C2H3O2)-1 (C2H3O2)-1 is the conjugate base of the acetic acid the Ka times the Kb of conjugate pairs = 1 X 10^-14 so Kb = (1 X 10^-14) / (1.8 X 10^-5) = 5.56 X 10^-10 that conjugate base reacts in water: (C2H3O2)-1 in water --> H(C2H3O2) & OH- Kb = [H(C2H3O2)] [OH-] / [(C2H3O2)-1] 5.56 X 10^-10 = [X] [X] / [ 0.8571 M] X2 = 4.76 X 10^-10 X = [OH-] = 2.18 X 10^-5 pOH = 4.66 since pH + pOH = 14 pH = 9.34