How do you write a balanced equation for the neutralization of acetic acid with NaOH?
474531
CH3COOH + NAOH =.CH3COONA +H2O
921837
NaHCO3 --> Na+ + HCO3-\nCa(NO3)2 --> Ca(2+) + 2NO3-\nAl2(CO3)3 --> 2Al(+3) + 3CO3 (-2)
825401
First of all I assume  Na2Co3 should be Na2CO3.\n\nFe(CO)5 + 4NaOH = Na2Fe(CO)4 + Na2CO3+ 2H2O\n\nFirst, balance out the metals (Na and Fe) then C, which was already balanced. Then I chose to balance out H and which automatically balanced out O.
894546
You did not put the volume of HCl to neutralize so we will do the neutralization to pH = 7 only.\n\nmoles of HCl = 35 0 g/l / 36.46 g/mole = 9.60 mol/l\nmoles of NaOH = 30 g/l / 40 g/mol = 0.75 mol/l\nOne mole of HCl will neutralize one mole of NaOH.\n\n9.60 * Vol(HCl) = 0.75 * V(NaOH).\n\nOnce you know if you have any NaOH left you can calculate how much NaOH to add to bring it to pH = 8 or 9\n\npH = 8 =====> pOH = 6  (pOH = 14 - pH)\nthen 6 = - log [OH]\nCalculate OH from last equation and this will give you the moles of NaOH you need to add to bring to pH = 8.  Same for pH = 9.
797533
Providing that the H3C6H5O7 is citric acid, that balanced equation is:\n\nH3C6H5O7 (s) + 3 NaOH (aq) --> Na3C6H5O7 (aq) + 3 H2O (l)\n\nThe net ionic equation is:\n\n3 H+ (aq) + 3 (OH-) (aq) --> 3 H2O (l)
250273
Enough to cancel out the H's. For the reaction with 2H's, add 2OH's to both sides. For the reaction with 3H's, ad 3OH's to both sides.\n\nH + OH --> H2O\n\nEDIT: Oh, well in that case, the H's cancel out, so you have just one H on the side that previously had 3. Then, just add one OH to both sides.
743363
1. Get the number of moles of carminic acid.\nsince 1 mole of NaOH neutralizes 1 mole carminic acid, number of moles NaOH equals number of moles carminic acid:  \n0.01802L * 0.0406mol/L = 7.3 x 10^-4 mol\n\n2. Determine the molar mass of carminic acid.\nSince you are given the weight of carminic acid, you can get the molar mass by simply dividing it with is number of moles from #1.\n0.3602g / 7.3 x 10^-4 mol = 492.34 g/mol\n\n3. Calculate for the molecular formula.\na) Assume 100g carminic acid such that weights of C, H, and O are as follows: C = 53.66g; H = 4.09g; and the remaining \nO = 42.25g\n\nb) Get the number of moles of each element by dividing the weights by atomic mass. \nC => 53.66g/ 12g/mol = 4.47mol\nH => 4.09g/ 1g/mol = 4.09mol\nO => 42.25g/ 16g/mol = 2.64mol\n\nThis is your empirical formula: C4.47H4.09O2.64\n\nc) To get the molecular formula, the molecular mass must be equal to 492.34 g/mol. In this case, you have to multiply the empirical formula by ~5 to get C22H20O13. To check,\nC = 22* 12g/mol = 264g/mol\nH = 20* 1g/mol = 20g/mol\nO = 13* 16g/mol = 208g/mol\nwhen you add all up, you get 492g/mol.\n\nSo, molecular formula for carminic acid is C22H20O13.
939677
Long answer:  Use the vol. and concentration of NaOH to figure out how many moles of NaOH you have which will = moles of OH (since NaOH has only one OH).  Formula is:\n\nMolarity = moles/liters (make sure the volume is in liters)\n\nIn a neutralization reaction the moles of OH = moles of H+\n\nOnce you have moles of H+, use that answer and the volume of HCl to calculate concentration of H+ which will be = to concentration of HCl since HCl has only one H+\n\nUse the same Molarity formula\n\nShort answer:  As long as the base (NaOH) and the acid (HCl) have only one OH and one H, the short form is:\n\nMb x Vb = Ma x Va\n\nwhere Mb = molarity of the base\nwhere Vb = volume of the base\nwhere Ma = molarity of the acid\nwhere Va = volume of the acid
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
877763
OK, I think I have this one pegged:\nThe original mixture is 10% acid and 90% water and the final mixture is 60% acid and 40% water after pouring some amount of the original mixture into 5 liters of pure 100% acid. Set up two equations:\nfor the acid: x (10) + 5 (100) = y (60) where x = original mixture volume and y = volume of resulting mixture.\nfor the water: x (90) + 0 (100) = y (40)\nthen these can be reduced to:\n10x + 500 = 60 y     and  90x = 40y\nsolve the simultaneous equations to find x and y\nThe volume of water in the final mix is equal to y times .40\nGood luck
905904
step 1 2NaOH + H2SO4 ---------> Na2SO4 + 2H2O\nstep 2 amount of NaOH = 0.2 x 25/1000= 5.0 x 10^-3 mol\nstep 3 amount of H2SO4 = s.0x10^-3 x 1/2 = 2.5 x 10^-3 mol\nstep 4 mass of H2SO4 = 2.5 x 10^-3 x 98 = 0.245g required\n\nNOTE\n^= power\nin step 4 it is multiplied by 98 because that is the molar mass of H2SO4 ie (2x1)+32+ (16 x 4) =98\n\ni do hope this is correct it has a been about 3 yrs since i did my A Level in chemistry
802376
YOUR ANSWER IS CORRECT!  Here's how i did it:\n\nYou can find the # mol of NaOH used:\n\nMolarity = (# mol)/(volume solution)\n0.4 = (# mol)/(.075 L)\n\nSolving, you get 0.03 mol of NaOH.\n\nUsing the balanced equation:\n\n3 NaOH + H3PO4 ==> Na3PO4 + 3 H2O \n\nYou see that 3 mol NaOH = 1 mol H3PO4.  Thus, if we have 0.03 mol NaOH, that means we have 0.01 mol H3PO4.\n\nNow we can solve for the molarity:\n\nMolarity = (0.01 mol H3PO4)/(0.05 L) = 0.2 M
708721
Here is the right equation for nitric acid and sodium bicarbonate\n\n\nHNO3 + NaHCO3 ---> NaNO3 + H2O + CO2\n\nIf you do not like the sodium:\n\nHNO3 + HCO3(-) --> NO3(-) + H2O + CO2
432315
You'll need to give the degree of dissociation of the acid CH3COOH. If it was a strong acid you have to used the formula \nMV/mv = 1   where M is molarity of NaOH,V is its volume\n            and m is molarity of CH3COOH, v is its volume\n\nActually the problem here is CH3COOH is a weak acid and it would not have dissociated fully, hence the above formula cannot be used.
955158
Find the moles of the acid CH3COOH and of the NaOH:\n\nn = C*V = 1*0.05 = 0.05 mol CH3COOH\n\nn = C*V = 1*0.05 = 0.05 mol NaOH\n\nAccording to the balanced chem. equation:\n\nCH3COOH + NaOH  --> CH3COONa + H2O\n\n1 mol of CH3COOH reacts with 1 mol of NaOH and gives 1 mol of H2O\n\nSo 0.05 mol CH3COOH reacts with 0.05 mol NaOH and gives 0.05 mol of H2O\n\nThat's the answer: 0.05 mol H2O
320061
Someone else has asked this same question before, and it really isn't that hard . . .\n\n60 ml of 0.250M HCl plus 60 ml of 0.125M NaOH will result in 120 ml of solution total.  All of the NaOH will have been neutralized by the HCl, leaving an excess of HCl, so the solution will be acidic -- the pH will be much less than 7.\n\n60 ml * (1.0 liter/1000 ml) * (0.250 mole/liter HCl) = .015 mole HCl\n\n60 ml * (1.0 liter/1000 ml) * (0.125 mole/liter NaOH) = .0075 moles NaOH\n\nThat means you will have 0.015 - 0.075 = 0.075 moles of HCl left over after the neutralization.\n\nI leave it for you to calculate the pH of a solution of 0.075 moles of HCl in 120 ml of solution.  Remember that pH = -log(base 10)[H+]
804535
Neutralization Reaction:\nHC2H3O2(aq) + KOH(aq) -->H2O(l) + KC2H3O2(aq)\nThe equation is already balanced.\n\na. False KC2H3O2(aq) and H2O are products.\nb. True\nc. False, some ions will cancel out but some will remain.\n\nThe ionic equation is as follows:\n(H^+)(aq) + (C2H3O2^-)(aq) + (K^+)(aq) + (OH^-)(aq) -->H2O(l) + (K^+)(aq) + (C2H3O2^-)(aq)\nRepeated ions are crossed off from both sides and the net ionic equation is:\n(H^+)(aq) + (OH^-)(aq) -->H2O(l)\n\nd. True\ne. False\n\nNote: the + and - signs were written with ^ to indicate that they were raised up (like a power).  Sorry if this notation is confusing.  Good luck and I hope that this helps!