If the Ka of a monoprotic weak acid is 1.9 × 10^-6, what is the pH of a 0.14 M solution of this acid?
We need to make an ICE table, but I don't know how. Just analyze this: The overall equation: HA + H2O <---------> A- + H3O+ Initial: HA = 0.14M A- = 0 H3O+ = 0 Change: HA = -x A- = +x H3O+ = +x Equilibrium: HA = 0.14M -x A- = +x H3O+ = +x Ka is 1.9 x 10^-6 Ka = products / reactants Ka = [A-][H3O+] / [HA] Ka = (x)(x) / (0.14 - x) 1.9 x 10^-6 = x^2 / (0.14 - x) 0.000000266 - 1.9x10^-6x = x^2 x^2 + 1.9x10^-6 - 0.000000266 (find x using quadratic formula) x = 5.148027533x10^-4 [A-]=[H3O+]=5.148027533x10^-4 pH = -log [H3O+] pH = -log (5.148027533x10^-4) pH = 3.2883591390407677760167701607 or 3.29
What is the ph of 5×10^-6 M Ca(OH) 2?
Assuming complete ionization Ca(OH)2Ca(OH) 2 is 5×10^-6 Min this case OH- concentration produce by water is not ignoredVidyanchal Academythanku
What is the pH of a solution with a hydrogen ion concentration of 10^-6M?
pH is a measure of H+ ion concentration→ the higher the H+ ion concentration, the lower the pH (i.e. closer to 0) and the more acidic the solution.The solution has a H+ ion concentration of 10^-6 MpH= - log (H+ concentration in moles ) = - log (10^(-6))= -1 x -6= 6Hence the pH of the solution is 6 , i.e. weakly acidic.
How do I calculate the pH of a 0.093 M NaF solution, where the Ka for HF = 7.1 x 10^-4?
The pH of your solution will be equal to 8.06.Sodium fluoride,NaF, is a soluble salt that dissociates completely in aqueous solution to give sodium cations, Na+, and fluoride anions, F−.NaF(s)→Na+(aq)+F−(aq)Since 1 mole of sodium fluoride produces 1 mole of fluoride anions, the concentration of the fluoride ions will be equal to that of the salt.[F−]=[NaF]=0.093 MThe fluoride ions will react with water to form hydrofluoric acid, a weak acid, and hydroxide ions, OH−, which is an indicator that the pH of the solution will be greater than 7.F−(aq)+H2O(l)⇌HF(aq)+OH−(aq)Final concentrations will be [F-] = [F-]-[HF] (or [OH-]) = 0.093-x, [HF] = x, [OH-] = xThe base dissociation constant, Kb, will be equal toKb=Kw/Ka=10−14/7.1⋅10^−4=1.41⋅10^−11This means that you'll getKb=[HF]⋅[OH−]/[F−]=x⋅x/(0.093−x) =x^2/0.093−xSince Kb is so small, you can approximate (0.093-x) with 0.093 to getKb=x^2/0.093=1.41⋅10−11Therefore, x=(0.093⋅1.41⋅10−11)^-2=1.15 x 10^−6You can now determine the pOH of the solutionpOH=−log([OH−])pOH=−log(1.15 x 10^−6)=5.94As a result, the solution's pH will bepH=14−pOH=14−5.94=8.06Better check my math but the basic approach is correct.
What is the pH of a saturated solution of Zn(OH) 2 (K_sp = 3.0×10^-16)?
First we need to write the equation for the dissolution of Zn(OH)2Zn(OH)2(s) → Zn 2+(aq) + 2OH-(aq)Then we write Ksp expressionKsp= [ Zn 2+][OH-]^2 notice that solids are never included in the K expression.Now, lets set up an ICE chart (solids are not a part of ICE chart either)Zn(OH)2(s) → Zn 2+(aq) + 2OH-(aq)I(initial) 0 0C(change) x 2xE(equilibrium)x 2xLet’s plug in the equilibrium concentrations into our Ksp equation from aboveKsp= x(2x)^2 = 3.0*10^-16X=4.2 *10^-6[OH-]=2x=8.4*10^-6pOH= -log[OH-] = 5.0757207pH = 14-pOH = 8.9243 = 9.0Hope this was helpful, and sorry for the formatting issues.Please visit, Chemistry Tutor NYC | Transformation Tutoring for more chemistry help!
An acid solution with pH=6 at 25℃ is diluted 10^2 times. What will the pH of the solution be?
The pH is 6.95REASON:here due to dilution we need to consider effect of H2O. And as it is becomes aqueous solution so it contains some amount of H2O hence its pH increases (less acidic).Therefore answer is :-Given concentration of H+ ion is 10^-6 and on dilution to 100 times it becomes 10^-8total concentration H+ = 1 x 10^-7+ 1 x 10^-8 =1.1 x 10^-7 M pH = 6.95pH = 6.95
How do you convert pKa to Ka?
The acid dissociation constant, Ka, is a quantitative indicator of the strength of an acid in a solution. According to the Bronsted-Lowry definition, an acid acts as a proton donor while a base acts as a proton receiver. Because Ka is often an extremely large number, chemists simplify it to a smaller number, called pKa, for convenience. Ka and pKa are the same thing expressed in different ways.we know that, pKa= -log KaHence, Ka[math]Ka[/math]=10^(-pKa[math]a[/math])Sample Problem : Find the Ka for the equation: H20 + HCl <==> H30+ + Cl- if the concentrations of H30+ and Cl- are both 0.1 Molars and the concentration of HCl is 10^-8.Ka= [H3O][Cl-]/[HCl]Ka= [0.1][0.1]/10^-8Ka=10^7Convert Ka to pKa.pKa = -log(10^7)pKa = -7Reverse the equation to convert pKa to Ka:Ka=antilog(-pKa)Ka=antilog(-7)=10^7I hope you find this useful! :)
A Carnot engine takes in 3,000 kcal of heat from a reservoir at 627 °C, and gives it to a sink at 27°C. What is the work done by the engine?
T1=273+627=900kT2=273+27=300N=W/Q1=1-T2/T1W/3000k cal=1–300/900W=2000kcal2000×4.2KJ=8.4×10^6 joule
What is the pH of 10^-3 mole of KOH in 100 liters of solution?
mole of KOH : 10^-3 moleVolume of solution : 100 LpH of solution : ….. ?Step 1. You have to know the difference between a strong base or a weak base.Potassium Hydroxide is a strong base because it ionizes completely in water.KOH → K+ + OH−Step 2. If you want to find out pH of the solution. You have to find out the concentration (Molarity) in solution.M : mol / LM of KOH : 10^-3 mol / 100 LKOH : 10^-5 mol/L or 10^-5 MStep 3. Find out the concentration of [OH-] in solution.KOH → K+ + OH-10^-5 mol → 1 . 10^-5 mol10^-5 mol → 1 x 10^-5 molKOH is a strong base and only have 1 valance OH- so that the concentration of OH- will have 10^-5 M.By using the formula of pH. You are going to find out the pH of potassium hydroxide in solution.[H+][OH-] : 1 x 10^-14[H+] 1 x 10^-5 : 1 x 10^-14[H+] : 1 x 10^-9And thenpH : -log [H+]pH : -log 1 x 10^-9pH : 9So, the pH of potassium hydroxide in solution is 9.I hope it will help you.