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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.
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 ...
A quantity in square brackets, [X], represents the concentration of the chemical substance X. It is understood that the symbol H + stands for the hydrated hydronium ion. K a is an acid dissociation constant. The Henderson–Hasselbalch equation can be applied to a polybasic acid only if its consecutive pK values differ by at least 3.
where denotes the natural logarithm, with the sum extending over all prime numbers p that are less than or equal to x. The second Chebyshev function ψ (x) is defined similarly, with the sum extending over all prime powers not exceeding x
The derivative of ln(x) is 1/x; this implies that ln(x) is the unique antiderivative of 1/x that has the value 0 for x = 1. It is this very simple formula that motivated to qualify as "natural" the natural logarithm; this is also one of the main reasons of the importance of the constant e .
When the acidic medium in question is a dilute aqueous solution, the is approximately equal to the pH value, which is a negative logarithm of the concentration of aqueous + in solution. The pH of a simple solution of an acid in water is determined by both K a {\displaystyle K_{{\ce {a}}}} and the acid concentration.
Pauling's second rule is that the value of the first pK a for acids of the formula XO m (OH) n depends primarily on the number of oxo groups m, and is approximately independent of the number of hydroxy groups n, and also of the central atom X. Approximate values of pK a are 8 for m = 0, 2 for m = 1, −3 for m = 2 and < −10 for m = 3. [28]
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).