Intermediate form diprotic acid

Triprotic Acids and Bases, and Beyond The treatment of a diprotic acid or base is easily extended to acids and bases having three or more acid-base sites. For a triprotic weak acid such as H3PO4, for example, we can treat H3PO4 as if it was a mono-protic weak acid, H2P04 and HP04 as if they were intermediate forms of diprotic weak acids, and P04 as if it was a monoprotic weak base. 

We treat HP04 as the intermediate form of a diprotic weak acid... 

To treat this case, we resort to the systematic treatment of equilibrium. The procedure is applied to leucine, whose intermediate form (HL) has no net charge. However, the results apply to the intermediate form of any diprotic acid, regardless of its charge. 

Equation 10-12 is a good one to keep in your head. It gives a pH of 6.04 for leucine, compared with pH = 6.06 from Equation 10-11. Equation 10-12 says that the pH cfthe intermediate form of a diprotic acid is close to midway between pK I and pK2, regardless of the formal concentration. 

The pH of the intermediate form of a diprotic acid is close to midway between the two pk, values and is almost independent of concentration. 

The method of successive approximations is a good way to deal with difficult equations that do not have simple solutions. For example, Equation 10-11 is not a good approximation when the concentration of the intermediate species of a diprotic acid is not close to F, the formal concentration. This situation arises when Kt and K2 are nearly equal and F is small. Consider a solution of 1.00 X 10 1M HM, the intermediate form of malic acid. 

Advice When faced with the intermediate form of a diprotic acid, use Equation 10-11 to calculate the pH. The answer should be close to 2 (p/C( + pK2). 

When alanine is dissolved in water, the pH of the solution, by definition, is the isoionic pH. Because alanine (HA) is the intermediate form of the diprotic acid, H2A+, [H+] is given by... 

In triprotic systems, there are two intermediate forms. The pH of each is found with an equation analogous to that for the intermediate form of a diprotic system. Triprotic systems also have one fully acidic... 

Consider HA-, the intermediate form of a diprotic acid. Kz for this species is 10-4 and Kb is 10-8. Nonetheless, the Kz and Kb reactions proceed to nearly the same extent when NaHA is dissolved in water. Explain. 

At the first equivalence point, B has been converted into BH+, the intermediate form of the diprotic acid, BH +. BH+ is both an acid and a base. From Section 10-1, we know that... 

Point C in Figure 11-4 shows where the intermediate form of a diprotic acid lies on a titration curve. This is the least-buffered point on the whole curve, because the pH changes most rapidly if small amounts of acid or base are added. There is a misconception that the intermediate form of a diprotic acid behaves as a buffer when, in fact, it is the worst choice for a buffer. 

Please note that equations (4.8-5) through (4.8-7), (4.8-12) through (4.8-14), (4.8-19) through (4.8-21), and (4.8-26) through (4.8-28) are identical when the concentration fractions are written merely as a2i a1) and a0j where a2 is the concentration fraction of the fully protonated form (H2A for a dipro -tic acid, HA+for a diprotic amino acid, H2B2+for a diprotic base), a0 that of the fully deprotonated form, while is the concentration fraction of the intermediate form. For the buffer strength of the solution of a diprotic acid and/ or its conjugate bases we then have... 

Point C (and C and C") is the first equivalence point. H2A has been converted into HA, the intermediate form of a diprotic acid. The pH is calculated from Equation 11-13 ... 

Draw the structure of the amino acid glutamine and satisfy yourself that it is the intermediate form of a diprotic system. Find the pH of 0.050 M glutamine. 

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