what is the pH of a .0025 M solution of KOH?
358982
NaOH is completely dissociated when it is dissolved in water.  A .0025M KOH solution therefore produces a [OH-] equal to .0025M.  The ion product of [H30+] and [OH-] is a constant, 1.0 x 10^-14 M^2.  By substitution, the [H30+] can be determined.  The pH can then be calculated.\n\n[H30+] = 1.0 x 10^-14 M^2/[OH-] = 1.0 x 10^-14 M^2/2.5 x 10^-3 \n\naaaand all that equals .4 x 10^-11... and you put that in correct scientific notation... 4 x 10^-12 M for the concentration of [H30+].\n\nWe then plug that in...   \n\npH= -log[H30+] = -log (4 x 10^-12) = 11.4\n\nThe pH of a .0025M solution of KOH is 11.4, which is greater than 7, making it a basic solution.\n\nStrong base at that.\n\nP.S.- If it is supposed to have the correct number of significant figures/digits, the answer would then be 11.
797573
pH 1.91 = .0123M\n\n     so, percent =  .0123M/.500M =.025 *100= 2.5%
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
323054
You sure did pick a really long question for explaining.  And I'll have to argue that all of the pH's do not have correct sig figs but the numbers are right.\n\n(Sorry about the numbers in the tables but this place doesn't seem to allow better tables)\n\nA) 25.0mL of HAc(acetic acid) and 10.0mL of NaOH\n\n*This is a reaction so we have to use moles*\n\nHAc : 25.0mL * (0.100mol/L) * (1L/1000mL)= 0.00250 moles\n\nNaOH : 10.0mL * (0.125mol/L) * (1L/1000mL) = 0.00125 moles\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00125      0.00250        0\nChange:   -0.00125     -0.00125      +0.00125 \nEquilib:      0         0.00125       0.00125\n\nHAc and Ac-(NaAc breaks apart) for an equilibrium\n*Use Molarity when no reaction is taking place (best shown by when no volume is changing)*\n\nBoth HAc and Ac- have same molarity:\n\n(0.00125 moles / 35mL ) * (1000mL/1L) = 0.0357M HAc and Ac-\n\nUsing the equilibrium equation:\n\nEquation:  HAc   <--> Ac-     +  H+         Ka = 1.8 * 10^-5\nInitial:   0.0357M     0.0357M  \nChange:    -x          +x         x\nEquilib:   0.0357-x    0.0357+x    x\n\nKa=1.8*10^-5 = [Ac-][H+]/[HAc] = [0.00357+x][x]/[0.00357-x]\nKa=1.8*10^-5 = [0.00357][x]/[0.0357] usually good approx.\nKa = 1.8 * 10-5 = x = [H+]\n\npH = -log [H+] = -log [1.8*10^-5] = 4.745 (3 sig figs as only after the decimal counts for logs)\n\nB) 25.0mL HAc and 20.0mL NaOH\n\n*See A for HAc moles and double the NaOH for this trial*\n\n*This is a reaction so we have to use moles*\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00250      0.00250        0\nChange:   -0.00250     -0.00250      +0.00250 \nEquilib:      0            0          0.00250\n\nCalculate molarity of Ac-:\n\n(0.0250moles /45mL) * (1000mL/1L) = 0.0556M Ac-\n\nUsing the other equilibrium equation (Kb):\n\nEquation:  Ac- + H20   <--> HAc     +  OH-         Initial:   0.0556M   -        0\nChange:    -x        -        +x        + x\nEquilib:   0.0556-x  -       +x         + x\n\nThis gives us the Kb equation\nKb = ([HAc][OH-])/[Ac-] = with approx ([x][x])/[0.0556]\nso Kb = x^2 /0.0556\n\n**but we need to calculate Kb\n**Kw = 1*10^-14 at STP I'll assume this for this problem\n**Kw = KaKb so Kb = Kw/Ka\n**Kb = (1*10^-14)/(1.8*10^-5)= 5.56*10-10 \n\nso x = 5.56 * 10^-6 = [OH-]\n\npOH = -log [OH-] = 5.253\nPH = 14 - pOH (at STP)= 8.747\n\nC)25.0mL HAc and 30.0mL NaOH\n\n*See A for HAc moles and triple the NaOH for this trial*\n\n*This is a reaction so we have to use moles*\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00375      0.00250        0\nChange:   -0.00250     -0.00250      +0.00250 \nEquilib:   0.00125            0       0.00250\n\nAfter the reaction there is some NaOH left over so it is the only thing that matters for the pH as it is a stong base.\n\nCalculate NaOH molarity\n(0.00125moles/55.0mL)*(1000mL/1L)= 0.0227M NaOH\n\nStrong Bases dissociate 100% so [OH-] = 0.0227M\n\npOH= -log[0.0227] = 1.644\npH = 14-pOH = 12.356\n\nThis should show how they calculted the half-neutralization point (A), the neutralization point (B) and after the neutralization point (C).\n\nIf you have questions on this, feel free to send me a message.  I taught this a couple weeks ago so I'm still pretty fresh on the subject.
173742
To determine pH for this mixture, one must know the molarity for the HCl, the molarity for the acetic acid, and the Ka for the acetic acid.  The goal is to find the concentration of the hydronium ion in solution.  Thus, for a strong acid, since it completely dissociates, the molarity equals the concentration of the hydronium ion.  But, it is more complicated for the acetic acid, which is weak.  We need to solve for the hydronium concentration, so we set up a Ka equilibrium equation, which with all the values added looks like:  Ka= (x^2)/ (Molarity of Acid).  Solve for x.  This gives you the hydronium ion concentration for the weak acid.  Add the two concentrations together and take the negative log of that number.\nRemember to use correct significant figures.  Also, it is not necessary to subtract x from the molarity of the acid in the equation, because since the Ka is small enough, when the math is done the percent ionization is less than 5%, so the x can be ignored without incurring significant errors.
592167
We assume NaOH is a strong base and therefore dissociates completely at 60 degrees C. a concentration of .002M solution of NaOH dissociating completely contributes .002 mols/liter of OH ions to the solution. Since pOH = -log[concentration of OH-], pOH = -log.002 = 2.699 (with correct logarithmic sig figs). pH is 14 - pOH, so the pH of the solution will be 14-2.699 = 11.301.
745984
To answer this question, you must know the Ka or Kb values for those acids or bases above.\nYou must then use those values to find the concentration of H+ (H30+ depending how you were taught it) or OH- ions.\nthese values can then be used to find pH or pOH depending which ion value you have.\nSimply perform the calculation \n\n-log[concentration H+ or OH-]\n\nthis will give you a pH value if you used H+ ions, or a pOH value if you used OH-. To get the pH from a pOH, you must take the value and subtract it from 14. (14-pOH = pH)
553835
Well to determine [H+] you use the equation that \npH=-log [H+] \nthus a solution with pH=3 will have a [H+] of 1x10^-3, \nand a solution with pH=8 will have a [H+] of 1x10^-8\nTo find [OH-] there are several equations you can use\n[OH-]=Kw/[H+], \nKw is the ionization constant of water which is equal to 1.0x10^-14\nThus you will have [OH-] of 1.0x10^-11 and 1.0x10^-6\nSo you answer will be that the solution with pH=3 will have a [H+]>[OH-]
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
817888
The number of moles of KOH used is:\n\n3.75 g / 56.1 g/mol = 0.0668 moles\n\nin 250 mL (1/4 L) this is 0.0688 * 4 = 0.267 M\n\nM1 * V1 = M2 * V2  or  M2 = (M1 * V1)/V2\n\n= (0.267 M * 27mL)/100mL = 7.209 / 100\n\n= 0.0721 M
853922
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
944217
all you do is break KOH into it's ions which makes the solution basic
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.
639844
[H3O+]=1 x 10^-4 => pH=-log[H+]=4 --- acidic\n[OH-]=1X10^-6 => pOH=-log[OH-]=6 => pH=14-6=8 --- basic\n\nHCl --> H+ + Cl-\n[H+]= 5.6 x 10^-4\npH = -log[H+] = -log[5.6 x 10^-4] = 3.25 --- acidic\npOH = 14 - 3.25 = 10.75\n-log[OH-] = 10.75 => [OH-]=1.78 x 10^-11\n\nSince HBr --> H+ + Br-\npH of 1 x 10^-4 of HBr = -log[1x10^-4] = 4 --- acidic\n\nSince Ca(OH)2 --> Ca2+ + 2OH-\npOH = -log[OH-] = -log(2 * 5 x 10^-4) = 3\npH = 14-3 = 11 --- basic
165248
(.002mol /20ml) (1000ml / 1L ) = .10 M NaOH\n\n\n pOH =  -log 0.10 =  1\n\n pH = 14.00 - pOH   = 14 - 1  = 13 = D
459258
start with computation of molarity of solution:\nM_BaOH = (51.24g/ MW BAOH)/ 1.2 L\n\nthen \nusing M1V1 = M2V2\nM_BaOH (x) = (0.1) (1L)\nsolve for x
224213
well, pH = -log[H+]. so if you just undo the log you will have the concentration of H+ (which is the same as H30+). then .. [H+} + [OH-] = 14 good luck!