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A strong acid, such as hydrochloric acid, at concentration 1 mol dm −3 has a pH of 0, while a strong alkali like sodium hydroxide, at the same concentration, has a pH of 14. Since pH is a logarithmic scale, a difference of one in pH is equivalent to a tenfold difference in hydrogen ion concentration.
With specific values for C a and K a this quadratic equation can be solved for x. Assuming [4] that pH = −log 10 [H +] the pH can be calculated as pH = −log 10 x. If the degree of dissociation is quite small, C a ≫ x and the expression simplifies to = and pH = 1 / 2 (pK a − log C a).
At half-neutralization the ratio [A −] / [HA] = 1; since log(1) = 0, the pH at half-neutralization is numerically equal to pK a. Conversely, when pH = pK a, the concentration of HA is equal to the concentration of A −. The buffer region extends over the approximate range pK a ± 2. Buffering is weak outside the range pK a ± 1.
The ocean contains a natural buffer system to maintain a pH between 8.1 and 8.3. [11] The oceans buffer system is known as the carbonate buffer system. [ 12 ] The carbonate buffer system is a series of reactions that uses carbonate as a buffer to convert C O 2 {\displaystyle \mathrm {CO_{2}} } into bicarbonate . [ 12 ]
The three species all have concentrations equal to 1 / K D at pH = pK 1, for which [Cr] = 4 / K D . [3] The three lines on this diagram meet at that point. Green line Chromate and hydrogen chromate have equal concentrations. Setting [CrO 2− 4] equal to [HCrO − 4] in eq. 1, [H +] = 1 / K 1 , or pH = log K 1. This ...
Because log(x) is the sum of the terms of the form log(1 + 2 −k) corresponding to those k for which the factor 1 + 2 −k was included in the product P, log(x) may be computed by simple addition, using a table of log(1 + 2 −k) for all k. Any base may be used for the logarithm table. [53]
The identities of logarithms can be used to approximate large numbers. Note that log b (a) + log b (c) = log b (ac), where a, b, and c are arbitrary constants. Suppose that one wants to approximate the 44th Mersenne prime, 2 32,582,657 −1. To get the base-10 logarithm, we would multiply 32,582,657 by log 10 (2), getting 9,808,357.09543 ...
The pH after the equivalence point depends on the concentration of the conjugate base of the weak acid and the strong base of the titrant. However, the base of the titrant is stronger than the conjugate base of the acid. Therefore, the pH in this region is controlled by the strong base. As such the pH can be found using the following: [1]