If a solution contains 0.01 mol HCl in 10L of solution, what's its molarity? Its H+ ion concentration? Its pH?
971190
Molarity is defined as number of moles in 1 liter.\nYou have 0.01 mol and 10 liters\nM = 0.01 / 10 mol/liter - 0.001 mol/liter = 0.001 M.  That's it.  You did it.\n\nNow pH = - log[H+].\n\nIf you write the equation\nHCl ---> H+ + Cl-\nThis means that you have the same molarity for HCl and H(+).  For both of them the coefficient in the chemical equation is equal to 1.  For other problems the coefficients may be different so always look at the balanced equation.\n\nso M(HCl) = M(H+) = 0.001.\n\nPutting this value into the pH equation:\n\npH = - log 0.001 = 3\n\npH + pOH = 14.\nCalculate the pH according to the pH definition and then use the above equation to find the pOH.
5197
More information is needed.  What is the concentration is it a percentage or a molarity.\n\nDo you need to find the concentration or just the amount of alchohal?
1108880
HCl   +  NaOH   ===   NaCl   +   H2O\n0.0015 ,  0.0016 ,             0        ,       0        at start of reaction (moles)\n0,           0.0001 ,            0.0015 ,    0.0015  at end of reaction(moles)\n\nevery mole of water contributes about 10^-7(ten to the power -7)\nmoles of OH-  and H+ . thus their contribution to the pH is neglected when there is a considerable amount of acid or base present  i.e more than 10^-5 moles/litre (not moles but CONCENTRATION)\nThus concentration of NaOH = 0.0001/90*1000 moles/litre{total volume = 50+40 ml} = 0.001 .\nas NaOH is a strong electrolyte like any other sodium compound  we can assume complete ionisation . thus[OH-]=[NaOH] = 0.001. now pH = -log[H+]=14-(-log[OH-])= 14+log[OH-]  =14+log[0.001]=11\nthus the pH=11
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
812718
A\n\nThe pOH of a solution is the negative log of the concentration of OH ions.  THis would be 9 in this case.  The pOH is the opposite of pH.  The scale is from 0 - 14.  The lower the pOH, the more basic the solution is.  pOH of 7 is neutral.  Lower than 7 is basic and higher than 7 is acidic.  If the pOH of a solution is 9,  the pH of the solution is 5.  The pOH and the pH should add up to be 14.  I hope this helps.
1104578
Use the equation M1V1 = M2V2\n\nM1 is the molarity of the HCL or 9.9 M\n\nV1 is the first volume of 102 ml but it needs to be changed to Liters. So it is 0.102L\n\nV2 is the total volume of the final solution so 525 ml (0.525 L) + V1 or 0.102L = 0.627\n\nM2 is what you are looking for - the final concentration or molarity.\n\nRearrange the equation to ((M1)(V1)) / (V2)  = M2\n\n(0.102L)(9.9M) / (0.627) = 1.61 M\n\nHope this helped!
808549
M=molesof solute/Liter of solution\n\nM= molarity
741598
Well first you need to find the Ka of tartic acid: 10^-pKa = 1.05 x 10^-3.\n\nThen you create and ICE Table and solve for the Equilibrium concentration of Hydronium ions which gives you 0.02 M\n\nLastly, you find the pH which is -log [H3O+] = 1.6\n\nHope this helps
909115
20 ml of gastric juice  of (........) molarity <-- acid \n29.8ml of NaOH of (.0210M) <-- base   \n\nMOLE of acid = MOLE of Base  \n\n.0298L  X .0210 M = 6.258 X10^ -4 mole of base.. \nnow,  u know that mole base = mole of acid only at equivalence point... \n\n6.258X10^ -4 moles of acid / .025 L ( this is what was used in begining)  = .025032 Molarity..  \n\nacid base problem .. memorize the units..   V = Liters, moles = moles, Molarity = moles/Liters ...    this problem they give u Liters, u need to know how much moles in beginning... to get Molarity
950979
pH = - log[H+], so [H+] = 10^(-pH)\n\nfor pH = 7, [H+] = 10^(-7)
422227
In your example, the concentration of the solution weight/weight would be (presuming the 100 ml is water, thus 100g) would be 10.7% (12 g solute/112 g total solution).
770958
1/1000 x molar mass (grams/litre) and solute in 1 litre
991589
pH = -lg [H+]\n\nHence, \npH = 3 means [H+] = 10^ -3\npH = 5 means [H+] = 10^ -5\n\npH = 3 has 100 times more H+ ions in the same volume of solution than that of a pH = 5, which means it is 100 times more acidic
887080
Okay first, you have to deduce how many moles of acid react with the base.\n\n1 mol HCl ----------> 1 Mol NaOH\n\n1 H+ --------> 1 OH-\n\nTherefore:\n\nMolarity Acid X Volume Acid = Molarity Base X Volume Base\n\nMaVa = MbVb\nMa(20cm3) = 1.2 X 25 cm3\nMa =  30/20\nMa = 1.5 moles per dm-3\n\nThat gives the MOLARITY of the acid. The concentration is in grams per dm-3. \n\n1.5 moles of HCl = (Molar Mass) X 1.5\n                            = 36.46 X 1.5\n                            = 54.69g\n\nSo effectively;\n\nThe Molarity = 1.5 moles dm-3\nThe Concentration = 54.69 grams per dm-3\n\nWhich ever you need!
101093
I don't even get your question here... \n\npH is a mesurment of the quantity of Hydrogen Molecules that is available in a solution.  The pH of a solution changes depending on the concentration of salt you put in.
832170
Molarity = moles of solute/liters of solution\n15.8 g of HF = 1.01 + 19 = 20.01 g/mol\n15.8 g HF / 20.01 g/mol HF = 0.790 mols of HF\n0.790 moles of HF / 0.200 L of solution = \n3.95M solution
853370
The concept of 1.5M hydrogen gas is rather odd, but, yes, you use the Nernst Equation.  Don't forget to convert the pH to a molarity of H+ ions.