what is the concentration of hydrochloric acid in the resulting solution?what is its pH?
952447
I think you probably mean that the half liter is mixed with 250 ml. Otherwise the volume afterwards couldn't be 750 ml. Okay, the amount of HCl in the half liter is\nn=V*c=0.5l*2.5 mole/l=1.25 mole (remember that M simply means mole per liter).\nThe amount of HCl in the 250 ml is\nn=V*c=0.25l*3.75 mole/l=0.9375 mole\nThe total amount of HCl is then 1.25 mole+0.9375 mole=2.1875 mole meaning that the concentration in the final solution is\nc=n/V=2.1875 mole/0.750 l=2.92 mole/L=2.92 M.\n\nSince HCl is a strong acid it is completely separated in H+ and Cl- in the solution meaning that\npH=-log(concentration of H+)=-log(concentration of HCl)=-log(2.92)=-0.5.\n(Strictly speaking this formula for pH is only valid in dillute solutions like 0.1 M and below, but it is the best estimate we can give without making this problem VERY compliated).
465978
45 cm3 = 45 mL\n\n45 mL * 1.2 M HCl = 54 mmol HCl\n\n60 mL * .9M HAc = 54 mmol HAc\n\nSince the material amount is the same, HAc (acetic acid, CH3COOH) will have a higher pH because HCl is a strong acid and it will have a low pH. CH3COOH is a weaker acid, so its pH would be higher. HCl is a strong acid because the H and the Cl are strongly solvated by water molecules because of the immense polarity of the H-Cl bond. CH3COOH is weak because, while the anion is resonance stabilized, you still have a molecule with a hydrophobic end and an electron donating group. (the CH3)
248635
To neutralize and  to have a pH of 7 you will need to add 200.g of NaOH \n\n Molar mass is 58.44g NaCl\n \nmass salt Concentration is 292.2g NaCl\n\nW% = 0.286%\n\nif, the M NaOH solu. is equal to  HCl solu.  of  .05M
250762
2 HCl + CaCO3 ----> CaCl2 + H2O + CO2
624341
A solution that contains an equivalent amount of an acid and its conjugate base.
904575
pH=-log(H+)\nthen H+ or H3O+ if done in water is:\n10^(-1*pH)\nFor plugging in 7.5 in above returns: 3.16 x 10 -8 M (molar concentration)\nsimilarly pH 9.6= 2.51 x 10 -10 M
264737
Dissolve the following in 800ml distilled H2O.\n8g of NaCl\n0.2g of KCl\n1.44g of Na2HPO4\n0.24g of KH2PO4\nAdjust pH to 7.4 or 7.6 with HCl.\nAdjust volume to 1L with additional distilled H2O.
1025655
Think of the net ionic equation which is occuring.\nSince both the HCl and Ca(OH)3 are dissolved in aqueous solution, broken into ions, just focus on the H+ and OH- part of it first.\n\nH+(aq) + OH-(aq) --> H2O(l)\nThe Hydronium ion and the Hydroxide ion react to form water.\nWhat is left over are the Ca+2 and Cl- ion which drop out of the net ionic eqaution since they form the very soluble compound, Calcium Chloride (CaCl2).\n\n2 HCl (aq) + Ca(OH)2 (aq) --> 2 H2O (l) + CaCl2 (aq)
355889
There are three species in attendance in this solution, H+ and ClO- ions (that come from the disassociation of hypochlorous acid) and hypochlorous acid itself. \n\nThe Ka value comes from the following equation: \nKa= [H+]*[ClO-]/[HClO]\n(the brackets stand for concentration)\n\nWe can set [H+] to the variable x, and because for each mole of H+ formed, one mole of ClO- is formed, [ClO-] is also x.  The concentration of HClO is initially .10M, and because x amount of HClO splits up into the two ions H+ and ClO-, the final concentration of HClO that we will be using is .10M-x.  \n\nThat gives us this following equation:  \nKa=x*x/(.10M-x) \nor \nKa=x^2/(.10M-x)\nand because Ka=3.5x10^-4, \n3.5x10^-4=x^2/(.10M-x)\n\nNow you can algebraically solve for the value of x, or use a graphing utility to do it for you.  \nCross-multiply:\n3.5x10^-4(.10M-x)=x^2\nNow distribute:\n3.5x10^-5 - 3.5x10^-4x = x^2\nand set it equal to zero:\n0=x^2 + 3.5x10^-4x - 3.5x10^-5\nand use the quadratic equation to solve for x:\nx= -3.5x10^-4 + sqrt((3.5x10^-4)^2 - (4*1*-3.5x10^-5)) all divided by 2*1\nx=.0057 \n\nSo the [H+] = [ClO-] = x = .0057 M.  \nThe HClO concentration is .10M-x=.094M.
581048
The coefficient before the aluminum is 2 and before hydrochloric acid is 6.  Therefore, there is a 2:6 ratio (or 1:3 if you reduce) between the reactants.  This means for every one mole of Al you will need three times as much HCl.  Thus, if you have 2.4 moles Al, you will need at least 7.2 moles of HCl.
533415
[H+] is the concentration of hydronium ion in mol/L.  Assuming HBr completely dissociates in water, it would 2.0 mol/L in this example - one mole of HBr provides one mole of H+.  If the dissociation is not complete, then you'd have to use the equilibrium constant to calculate [H+].
625417
pH= -log(6.0 x10^-3)\npH is about 2.22
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
698593
You are not giving us all of the info.\nYes the total or initial concentration of the acid is C=(8.345*10^-3)/0.05 = 0.1669 M\n\nFor a monoprotic acid Ka=[H+][A-]/{HA]\n[A-]=[H+]=10^-pH=10^-4\n[HA]=C-[H+]= 0.1669-0.0001= 0.1668\nso you have calcualted Ka correctly.\n\nKa depends on the nature of the acid and temperature (for small molecules whereas for bigger molecules like proteins it is also affected by the nature of the surrounding chemical groups).\n \nThus addition of base doesn't affect Ka, only the pH.\nIs the base monoprotic or not? If not, what's the valency? Is it strong or weak? If it's weak what's the Kb? Is it the conjugate base of the weak acid? Without these info we can't calculate the pH.
386784
Well first of all you have to specify if the H's you want to account for are either comming from HCl alone or both HCl and some H's comming from the dissociation of water.\n\nI suppose that only the number of moles of H's comming from HCl is really worth while working the question for, since the H's of the dissociated water does not change the pH of the 0.10M solution.\n\nTherefore:\n\n0.10moles are in 1000mL\n? moles are in 50mL\n\n50 x 0.1 / 1000 = 0.005moles\n\nTherefore we can say that 0.005moles of HCl in 50mL of water results in a concentration of 0.10M
1014284
2.70\n\nregard what ppl saying below, I can make this a little bit more clear.\n\nFirst of all, since the concentration of HCl is 0.00200M and HCl is a strong acid, so we can safely assume 100% dissociation. So, we have 0.02M hydronium ions [one to one ratio]; even if we consider the dissociation of water, it will not make any difference. the dissocation of water is depressed by the presence of "high" concentration of hydronium ions. \n\nyou can even calculate how much water actually dissociates into hydroxide and hydronium ions using the disociation constant of water, 10^-14. But I can guarantee you this will not make ANY difference in pH, say less than 1% for sure.\n\ndissociation of water need to be put into consideration ONLY IF the concentration of hydronium ions is VERY low, say 10^-6 -10^-7, or pH around 7, then it will make a difference. \n\nSo, pH = -log[H3O+]= -log[0.002]= 2.70
295875
for each one you are given a ratio i am assuming you are combining them both, make sure you write out a balanced equation for each one to make sure, some may not be a 1 to 1 ratio, next Molarity=mol/L therfore 1M = 1 mol per liter, use your periodic table to figure out the molecular weight for each one, each one of these is a strong acid, strong base reaction, therefore they will create water as a product and a salt, in the first one you are given 1.0 mol of HCl which is reacting w/ NaOH 1 to 1, therefore you only need to convert to # of moles of NaOH i am assuming you can do this hopefully this helps