pH level of 0.5M Nitrous Acid(HNO2,Ka=4.0 * 10^-14)?
929685
When you have weak electrolytes, (equillibria in general) you should make the following table.\n\n1. write the reaction\n\nHNO2 <=> H+ + NO2-\n\n2.Put the initial amounts (expressed as concentration) of each molecule. In this case we have only the concentration of HNO2, so\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\n\n3.Assume that an unknown amount "x" of the reactants react.\nThen write also how much of each product you expect to be produced according to the stoichiometry. In this case 1 mole of HNO2 gives one mole of H+  and NO2-, so x mole/L HNO2 will  give x mole/L of each and our table becomes\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\n\n4. Ka is an expression for the concentrations at equilibrium. So let's see what are those. For the reactants you will have initial-what reacted. We have only HNO2 with 0.5-x. For the products it will be the initial+what was produced. We didn't have any intial so it is only what was produced (x for both of the products). Thus the table in its final form is \n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\nAt Equil. .. .. 0.5-x .. .. .. x .. .. .x\n\nKa= [H+][NO2-]/[HNO2] = x^2/(0.5-x)\nNow you either solve the quadratic equation and find x, or we assume that 0.5 >>x and practically 0.5-x=0.5.\nThen Ka=x^2/0.5=4*10^-14 =>\nx=squareroot(2*10^-14)= 1.4*10^-7. It is much smaller than 0.5 so our assumption is valid (otherwise we would have to solve the quadratic and get an exact solution).\n\nFor a simple level of chemistry we would now say\npH=-log[H+]= -logx =-log(1.4*10^-7) = 6.85\n\nHowever for more advanced and accurate expectations we see that the H+ coming from HNO2 is found to be in the same order of magnitude as that coming from the self dissociation of water (10^-7 for distilled water). Therefore in this case we cannot ignore the self dissociation of water (as we usually do) for calculating the pH. \nThere are 2 approaches for this case, here is one of them.\nRemember that we didn't put an initial amount for H+ since in most cases we ignore the self dissociation of water. Now we will consider that we had pure water, so the initial concentration of H+ is 10^-7 and we want to see how much the HNO2 will dissociate under these conditions so our table becomes\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5 .. .. .. 10^-7\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\nAt Equil. .. .. 0.5-x .. 10^-7+x. .x\n\nand Ka=(10^-7+x)x/(0.5-x)\nNow we have to solve the quadratic so we can't do any simplification\n\n0.5Ka-Kax=10^-7x +x^2 =>\nx^2+ (10^-7+Ka)x-0.5Ka=0\n\nThe only acceptable solution is the positive one\nx=0.5(- (10^-7+Ka)+ squareroot (D))\nwhere D=(10^-7+Ka)^2-4*(-0.5Ka)=\n=(10^-7+4*10^-14)^2-4*(-0.5*4*10^-14)\n=9*10^-14\n\nand x=0.5*(-((10^-7))+4*10^-14) +3*10^-7) =10^-7\n\nso now pH=-log(10^-7+x) =-log(2*10^-7) = 6.70
541583
pH from 0 to 7: Acid solution.\nPH from 7 to 14: Basic solution.
907196
The concentration of H+ is 3.0 x 10^-4, so you know that the pH will be around 4.  What you need is a calculator that has a LOG button to convert the 0.0003 to an integer value.\n\nSo on your calculator you press the LOG button and then type in 0.0003 and then a ")" and press ENTER.  \n\nYou should get a result that says: -3.522878745 or so.  You need to round off to -3.5 and since pH is the NEGATIVE log of the hydrogen ion concentration you get rid of the minus.\n\nThe answer is 3.5.\n\n1x10^-3.5 is approximately equal to 0.0003.
699263
V=nRT/P\nn(O2)=3/2(0.0022mol of KClO3) =0.0033 gmol\nR=0.0826\nT=294 k\np = p (O2) + p (water vapour)\np(O2) = 0.926 Atm\nV=(0.0033)(0.0826)(294)/0.926=0.087 liter = 87 ml
1073643
the easiest way is to use a pH paper to test the pH. and just compare the reading with the one given with the test kit. \n\nOr you may do some neutralisation. And then calculate with the  amount of alkali you use for neutralisation. By that you can calculate the concentration of hydroxide ions in the alkali that you use. Then calculate the pOH for the alkali you use. pH=14 - pOH.\neg. \n1mol of NaOH gives you 1 mole of OH ions.\nso the pOH = -log1 = 0\npH = 14 - 0 = 14\n\nJust some rough guide. Hopefully it helps...Good luck.
744922
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
715311
HNO3 (nitric acid) is a strong acid, which means that when HNO3 is dissolved in water, essentially all the HNO3 molecules dissociate into hydrogen ions and nitrate ions:\n\nHNO3(aq) -> (H+) + (NO3-)\n\nEach mole of HNO3 that reacts produces 1 mole of H+ ions.\n\npH is defined (approximately) as:\n\npH = - log[H+]\n\nwhere [H+] means the molar concentration of hydrogen ions, and "log" means the base-10 logarithm.\n\nFor this problem, then,\n\npH = - log(2.5*10^-5) = 4.60\n\n[Strictly speaking, the pH is defined as the negative logarithm of the thermodynamic activity of the hydrogen ion, not the concentration.  For dilute solutions, though, the activity (which is a unitless quantity) is approximately equal to the numerical value of the molar concentration.  This is not something that is generally dealt with in introductory chemistry classes, though.]
502184
5H2O2(aq)+2KMnO4(aq)+\n3H2SO4(aq)-->\n2MnSO4(aq)+K2SO4(aq)+\n8H2O(l)+5O2(g)
834164
Mg(OH)2  --->  Mg^2+  +  2OH^-\n\nk(sp) = [Mg^2+][OH^-]^2 \n\nIf pH = 12.8, then\npOH = 14 -12.8 = 1.2\ntherefore, [OH^-] = 10^1.2 =  0.0631 (ask your calculator!)\nnow, k(sp) = 1.2 * 10^-11 = [Mg^2+] [0.0631]^2 \nIf you solve for [Mg^2+] , you will have the answer.\n\nIf you need more help, please email me or re post.
912510
That is the molecular formula for nitrous acid, written HNO2.
832121
Nope, :s sorry\n\nIts the same as before. The K(w) of water is 1.0 *10^-14\nK(w) = [H+][OH-]\n1.0*10^-14=[H+][2.8*10^-4]\n[H+]= 3.57*10^-11\n\npH=-log[H+]\n the pH is 10.4\n\nand pOH =-log[OH-] u have that in the ques\n\nHappy trails
825261
First you must balance the equation:\n4NH3 + 5O2  -> 4NO + 6H2O\nthere are 4N, 12H and 10O on each side\nyou have 4.78 moles of O2 you need to divide it by the number of moles of O2 that react and multiply it by the number of moles of NO produced.\n4.78 / 5 x 4 = 3.824moles\nHope this helps
639146
It will be the pH of the sodium hypochlorite formed.  Determine that amount, look up the Ka of the acid and solve.
948423
You are going the wrong way.\nThe dissociation constant of a weak monoprotic acid is Ka. If you have a polyprotic acid then you have Ka1,Ka2,etc and more than one equivalence points.\n\nDo the plot, find the equivalence point, and then pKa=pH for 1/2 volume of base of that needed to reach the equivalence point.\nSince you have a weak acid you don't have the equivalence point at pH=7 but at higher pH (The equivalence point will be at the middle of the rising curve). \nThen Ka=10^-pKa\n\nYou can have a look at \n\nhttp://www.shodor.org/unchem/basic/ab/\n\nhttp://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html
756159
n   = pV/RT = 1.03*1.02/284*0.0821 = 0.045moles\n\nmoles of Nitrogen = 0.045*0.78=0.0351\n\nmoles of oxygen = 0.022*0.045 = 0.0098mole\n\nweight of the gases = 0.0098*32 + 0.0351*28.02 =0.316+ 0.986=1.3g\n\ndensity 1.3 g/l
377149
Nitrogen (N2) 78.084 % by volume\nOxygen(O2)20.9476 % by volume\nArgon(Ar)0.934 % by volume\nCarbon Dioxide(CO2)0.0314 % by volume
670103
Assuming that NO2 is an ideal gas is these conditions (which is a reasonable assumption at such a low P), the number of moles into that flask is: n = PV/(RT) where P = 432 mmHg = 0.568 atm, V = 0.5 L, T = 60 C = 333 K, R = 0.082 L atm/(mol K).\nThen: n = 0.01 mol and, taking into account the Avogadro Number (in number of particles/mol) the number of molecules is NM = 0.01*6.022x10^23 = 6.022x10^21.