What is the [H3O+] in a solution that consists of 1.5 M NH3, and 2.5 M NH4Cl? Kb= 1.8 x 10-5?
pOH = 4.7 + log 2.5/1.5 =4.85 pH = 9.15 [H3=+]= 10^-9.15 = 9.2 x 10^-10 M
1. (5) The pH of a 2.0 x 10–5 M calcium hydroxide solution is ______. 2. (5) A 0.10 M NH3 sol?
1. (5) The pH of a 2.0 x 10–5 M calcium hydroxide solution is ______. 2. (5) A 0.10 M NH3 solution is 1.3% ionized. Calculate the H+ ion concentration. NH3 + H2O !" NH4+ + OH– Can you show the work so I can understand it?? Thanks so much!!!
Calculate the pH of a 0.20 M solution of NH4Br?
NH4Br is a acidic salt. The NH4 ion will react with water (hydrolysis) and form NH3 and H3O + (hydronium ion). To solve for pH, you must first solve for [H3O]. To do this, you first need to find the Ka for NH4 +. The Kb for NH3 is 1.8 x 10-5, the Ka is 5.6 x 10-10( 1.0 x 10-14/1.8 x 10-5). The Kb expression for the above reaction is: Kb = [NH3][H3O+] / [NH4+] 5.6 x 10-10 = x squared/.20M x = [H3O+] = 1.06 x 10-5 pH = -log [H3O+] = -log [1.06 x 10-5] = 4.97
Handerson-hasselbalch equation buffered solution question?
NH3 + H2O <-----> NH4+ + OH- pKb = -log(1.8e-5) pkb = 4.74 pKa= 14 -4.74 pKa = 9.25 mols of NH3 = 0.32 * 2.0 = 0.64mols NH3 mols of NH4+ = 0.26 * 2.0 = 0.52 mols NH4 0.10mol HCl - 0.64mols NH3 = 0.54mols NH3 0.10mols HCl + 0.52mols NH4+ = 0.62 mols NH4 pH = pka + log ( [Base]/[Acid]) pH = 9.25 + log( 0.54 / 0.62) pH = 9.19
Calculate the pH of a 5.0 M solution of aniline (C6H5NH2, Kb = 3.8 x 10-10)?
We need the Ka value: Kb*Ka=1x10^-14 (1x10^-14)/(3.8x10^-10)=2.63*10^-5 (Ka) We also need the ICE chart: Initial: C6H5NH2(5.0M) C6H5NH1(0.0M) H(0.0M) Change: C6H5NH2(-x) C6H5NH1(+x) H(+x) Equilib.: C6H5NH2(5.0-x) C6H5NH1(0+x) H(0+x) And the equilibrium constant expression: [C6H5NH2]/([C6H5NH1]*[H]) Now plug the values from the ICE chart in: ([0+x]*[0+x])/[5.0-x] Brackets only signify concentration, so to make it less complicated we can get rid of them. That equation will equal Ka, so we write it as 2.63*10^-5 = ((0+x)*(0+x))/(5.0-x) Use a calculator since it's a lot faster than by hand, which gives: x=.011 Refer back to the ICE chart, so that the concentration of H is (0+.011) molarity -log(.011) = 1.96 pH Round to 2 sig figs: pH = 2.0
What is the pH of a solution by mixing 50 ml of 0.4N HCL and 50 ml 0.2 N NaOH?
Mole = concentration x volumemole of NaOH = 50ml x 0.2 = 10.0mole of HCl = 50ml x 0.4= 20.010 mole of NaOH exactly neutralized by 10 mole of HCl.NaOH + HCl = NaCl + H2Othat is 20 -10 = 10 HCl moleCon of H+ = no of mole of H+ / total volume of solution (50+50 ml = 100 ml)= 10/100 = 0.1pH = log ( H+)= log (0.1) = 1s0 the pH of solution will be one(1) .
Calculate the [H3O+] of a 0.10 M solution of NH4Cl in H2O at 25°C (Kb for NH3 = 1.8 x 10-5)?
Hi Rst5566, I am french (Boulogne sur mer 62200 - FRANCE) NH4(+) is an acid NH4(+) + H2O <------> NH3 + H3O(+) Ka = [NH3] x [H3O(+)] / [NH4(+)] [NH3] = [H3O(+)] [NH4(+)] = C – [H3O(+)] . . . then : Ka = [H3O(+)] ² / (C – [H3O(+)] ) Ka = Kw / Kb = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.6 x 10^-10 . . . that give : [H3O(+)] ² + 5.6•10^-10 x [H3O(+)] - 5.6•10^-10^-11 = 0 The resolution of the second degree equation gives : [H3O(+)] = 7,5•10^-6 mole/Liter . . . pH = -log [H3O(+)] = 5,13 I hope to have answered your question. .