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pipette out 8.5 ml perchloric acid in to 500 ml acetic acid and add 21 ml of acetic anhydride make up to volume to 1000 ml with acetic acid.Stand iday this solution and and check the water content not exceedsto 0.025 to 0.5% then standardize the solution by PHP

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It depends on just how dilute it is. If we're talking about a strong acid - in this case, acids like perchloric (HClO4), sulfuric (H2SO4), or nitric (HNO3) - and assume that the acid dissociates completely (an ideal situation; in sulfuric acid's case, only the first proton is assumed to dissociate), the pH of the acid is the negative logarithm of its molarity. For example, if you have 1 × 10-3 M (.001 mol · L-1) hydrochloric acid (HCl), its pH will be -log(1 × 10-3) = 3. (Since real life is a non-ideal situation it will actually be slightly higher, but we can disregard that.) That's the easy part and only applies to strong acids. For other (weak) acids of formula HA ⇌ H+ + A-, the pH is dependent upon the acid dissociation constant pKa, in which case pH = pKa + log([A-]/[HA]). Say you have a weak acid with a pKa of 2.0 and a molarity of .01 M. Since pKa = -log(Ka), that means that Ka = .01. The definition of Ka is [A-][H+]/[HA]. Let's call [A-] and [H+] x for this purpose; this makes [HA] = .01 - x; thus, .01 = x2/(.01 - x). Solving for x gives x2 + .01x - .0001 = 0; using the quadratic formula we get .00618 M. Now we may derive the pH. pH = pKa + log([A-]/[HA]) = 2 + log(.00618/.00382) = 2.21.


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