there is not
pKa+pKb=pKw=14
I will assume you are asking about the pH of pure water if pKw is 14.26. The relationship between pH, pOH, and pKw is as follows: pH + pOH = pKw. If it is pure, neutral water (no acids or bases present), then pH = pOH, so: pH + pOH = 14.26 2(pH) = 14.26 pH = pOH = 7.13
red car = rotes Auto red car = roter Wagen red car = roter PKW
The address of the On The Chisholm Trail Association is: 1000 Chisholm Trail Pkw, Duncan, OK 73533-1539
http://www.alfalaval.com/campaigns/tankequipment/mixing-and-agitation/product-overview/pages/product-overview.aspx?pkw=industrial%20mixing%20equipment is a website that you can find an industrial mixing equipment.
This is a weak base problem. Assume F = 0.028M Ka = 4.9 10^-10 Kb = Kw / Ka Kw - 10^-14 therefor pKw=14 x = [OH-] Kw / Ka = Kb = x^2 / (F-x) solve for "x" by use of quadratic formula to get the [OH-] pOH = -log [OH-] then plug pOH into the pH equation pH =pKw - pOH
No it is not true because value of pkw changes with change of temperature. It's value is 14 at 250C and less at high temperature
PKa = -log Ka so if you multiply across by -1 and then taking the antilog you can get Ka Ka.Kb = Kw where Kw = 1.0 x 10^14 PKa + PKb = PKw = 14 that should give you a start.
The concentration of hydroxide ion is realted to pH by the pKw (10-14) At pH 9 the concentration of OH- is 10-5, at pH 3, 10-11. The ratio is 106 so there are a million times as many OH- in pH 9.
The concentration of hydroxide ion is realted to pH by the pKw (10-14) At pH 9 the concentration of OH- is 10-5, at pH 3, 10-11. The ratio is 106 so there are a million times as many OH- in pH 9.
The concentration of hydroxide ion is realted to pH by the pKw (10-14) At pH 9 the concentration of OH- is 10-5, at pH 3, 10-11. The ratio is 106 so there are a million times as many OH- in pH 9.
Because water molecules can both split off and accept a proton H+ :H2O + H+ ---> H3O+ (hydroxonium ion)H2O ---> H+ + OH- (hydroxide ion)This is called the 'Water equilibrium of internal protolysis: 2H2O OH- + H3O+ pKw = 14 or Kw = [H3O+]*[OH-] = 10^(-14)