suppose the pH changes from 4 to 8, what is the change in H+ concentration?
952799
"p" is defined as minus the log...pH is saying -log(H).\n\nTherefore, -log(4) = 1 x 10 to the minus 4 = .0001\n\na pH of 8 = 1 x 10 to the minus 8 = .00000001\n\nso the change in concentration is 10 to the 4 or 10,000 times
271476
pH = 4.75\n\ntechnically, the pH can be anything.  But, for the most effective acetic acid buffer, the pH would be 4.75 because the pKa of acetic acid is 4.75.  When the pH = pKa, the amount of acid and base are equal, thus its resistant to pH changes is greatest.
1057594
I think this is how you do it:\nacid and base is measured in terms of pH and pOH. as you know, the H makes it acidic and the OH makes it basic, and if you mix an H and OH, you get water, which is to say it was neutralized.\nSo, let's say you have HBr with a pH of 4. You that pH is a measure of the concentration of H present in the mixture. So, to get the actual number you do -10ph, and you'll get a number like 10^-5 or something. So, to neutralize that many Hs, that's how many OH's you need. so, you'll need a pOH 4, which is the same thing is a pH of 10 (pOH = 14 - pH). \nhope that helps, good luck. and read the book, and do some of hte exercised in the book.
397908
The different acids have a different pH so turn a differetn colour.\nThe concentrations are different. The HCl is more concentrated as it will fully dissociate.
731164
pH: pH is a measure of the acidity or hydrogen ion concentration of a solution. It is measured on a scale of 0-14 with pH values below 7 being acidic, values above 7 being basic and a value around 7 is neutral. As the pH drops into the acidic range an enzyme tends to gain hydrogen ions from the solution. As the pH moves into the basic range the enzyme tends to lose hydrogen ions to the solution. In both cases the changes produced in the chemical bonds of the enzyme molecule result in a change in conformation that decreases enzyme activity.\n\nHope this helps :)
355477
Buffers resist change in pH. This is generally a result of the buffer containing a mix of a weak acid and the salt of the conjugate base. \n\nWhen you add NaOH to the buffer, it dissociates into Na+ and OH-. The OH- reacts with the H+ ions which the weak acid has dissociated into (along with the conjugate base) up to the acid dissociation constant, Ka, of the acid.\n\nThis shifts the equilibrium of the dissociation reaction of the acid, and more dissociates, maintaining a relatively stable pH, as compared to when the NaOH is added to the distilled water. (If you had instead added a base, it would instead have shifted the equilibrium of the base, and, instead, more of the base would have reacted with the free H+, again, keeping the pH relatively constant)
272078
I take it you mean the concentration of hydronium ions (H3O+) at a pH of 6.2. ok, the equation you need is this:\npH = -log [H+]\n\nTo solve, do this:\npH = -log [H+]\n-pH = log [H+]\n10 ^ (-pH) = [H+]   (if you don't understand this step, look up logarithms. Basically, a= log b means exactly the same thing as 10^(a)=b )\n\n10^(-6.2) = [H+]\nYou have 10^(-6.2) mol/ L of hydronium ions.\n\nAnother useful equation is this:\n[H30+] * [OH-] = 10^-14\nwhich will give you the concentration of OH- ions when you have the concentration of hydronium ions, or vice versa.
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
763383
a) neutral\nb) basic\nc) basic\nd) acidic\ne) acidic if in solution\nf) basic.\n\npH is a scale ranging from 0-14.  7 is the intermediate and a pH of 7 means that the solution is neutral.  If the pH is below 7, then there are more H+ than OH- in the solution.  Hence the solution becomes acidic.  If the pH however is greater than 7, then there is less H+ than OH- in the solution and the solution is basic.\n\nFor the acid H2SO4, when in solution, it will dissociate to form H+ and SO4-.  The H+ gives the acid its acidic property.  Similarly when Ca(OH)2 is dissolved, it dissociates into Ca2+ and OH-.  The OH- gives the compound its basic property.
887006
pH= -log([H30+])\n\n\njust put the #'s in the above eqn.\n\nfor pOH, its pOH= -log([OH])\nand for pH given [OH], its pH= 14 - (-log([OH])
240687
4m+H-(4m-h) =\n4m+H-4m+h=\n4m-4m+H+h=\n0+H+h=\nH+h
455471
pH (potential hydrogen) is a measure of the activity of hydrogen ions (H+) in a solution and, therefore, its acidity or alkalinity. In aqueous systems, the hydrogen ion activity is dictated by the dissociation constant of water (Kw = 1.011 × 10−14 at 25 °C) and interactions with other ions in solution. Due to this dissociation constant a neutral solution (hydrogen ion activity equals hydroxide ion activity) has a pH of approximately 7. Aqueous solutions with pH values lower than 7 are considered acidic, while pH values higher than 7 are considered alkaline..\n\npH = -log(concentration of H+)\njust use the equation to figure it out.\nhttp://en.wikipedia.org/wiki/PH
664234
NH4ClO4 --> NH4^+ + ClO4^-\n\nNow in a water solution,\nClO4^- + H2O --> HClO4 + OH- --> This doesn't happen because HClO4 is a strong acid.  Therefore the only contribution to the pH is from the NH4^+\n\nNH4^+ <=> NH3 + H^+\nYou start with [NH4^+] = 0.40 M, and [NH3]=[H^+] = 0 M; after undergoing equilibrium [NH4^+] = 0.40 - x, while [NH3]=[H^+] = x; Ka = [NH3][H^+]/[NH4^+] = (x)(x)/(0.40-x) = Kw/Kb = (1 x 10^-14)/(1.8 x 10^5)\nAssume x << 0.40, so x^2/0.40 = 5.56 x 10^-10\nx = 1.49 x 10^-5 M = [H^+]\npH = -log(1.40 x 10^-5) = 4.827
599859
It can vary, based on the conditions. However, according to Harris's Quantitative Analysis (2002), it's 4.69*10^-11\n\nLet's be clear. That is the equlibrium constant for the reaction whereby the hydrogen carbonate loses a proton:\n\nHCO3- --> H+ + CO3-2
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
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
427689
seems like a homework question so use the equation pH=pKa + log [base]/[acid].  As a check for your answer when [base]=[acid]  the pH will equal the pKa.  the acid and base are  close in concentration so your answer will be around a pH=9.  pOH = 14-pH will help you on the other question.