A) pH = 2.9. In order to calculate pH, you must know the [H+] concentration. The Ka expression will allow you to determine the [H+] if you know the concentration of HCH02. In order to solve this problem, you need to determine the molarity of the formic acid and then use this value to calculate the [H+] and, ultimately, the pH. The calculation proceeds as follows [Pg.345]

Indicators can be used to estimate the pH values of solutions. To determine the pH of a 0.01 M weak acid (HX) solution, a few drops of three different indicators are added to separate portions of 0.01 M HX. The resulting colors of the HX solution are summarized in the last column of the accompanying table. What is the approximate pH of the 0.01 M HX solution What is the approximate Ka value for HX [Pg.340]

Just as the K p values of low-solubility solids are listed in handbooks, so are the Ka values of most weak acids. They can be used to determine [H+] and the pH of solutions of those acids. If the ionization of the acid is the only source of H" and A ions, we can start with the acid constant equation, multiply both sides by [HA], and substitute [H" ] for its equal [A ] [Pg.554]

Thus, the equations of isotherms (30) and (31) can be used to determine the isoelectric points of proteins by the dependence of their sorption on pH. This approach is realized in Refs. [41,42]. However, it should be noted that this determination is only possible for a low affinity of protein molecules to surface sites. As Figs. 12 and 13 show, for a high affinity (a low Ka) a significant sorption is observed also at pH>pI for cation exchangers and at pH

During titration up to the first equivalence point, an HA /HaA buffer region is established. At the first equivalence point, a solution of HA exists, and [H ] /Ka Ka2- Beyond this point, an A /HA" buffer exists and finally at the second equivalence point, the pH is determined from the hydrolysis of P. If the salt, A , is not too strong a base, then the approximate equation given can be used to calculate [OH ]. Otherwise the quadratic equation must be used to solve Equation 7.29 (see Example 7.19). [Pg.282]

Table 4 summarizes some experimentally determined Ka values for a variety of apatites and contaminants. Some values are veiy large (> 10,000), indicating the propensity for apatites to sorb and/or induce surface precipitation reactions. These values are useful in terms of understanding the broad affinity that apatites have for various solutes. However, they reflect the operationally defined conditions particular to each partition study (pH, I, solid-to-liquid ratio, and apatite type). [Pg.446]

If you recall, we were using a buffer that was composed from 0.30 mol HC2H302 and 0.30 mol NaC2H3Oz dissolved in 1.0 liters of solution. The Ka of acetic acid is 1.7 X 10-5. Using the Henderson-Hasselbalch equation to determine the pH, we get [Pg.337]

Since buffers are an example of the common-ion effect, the concentration of ions (hence the pH) of a buffer solution can be calculated using a method that is similar to the method you used to determine solubility in a common ion effect problem. You will use Ka, however, instead of Kgp. Also, buffers are in homogeneous equilibria, unlike saturated solubility systems, which are in heterogeneous equilibria. Therefore, you need to consider the initial concentration of the reactants. (The reactants are not solids, so their concentration is not constant.) [Pg.440]

Hydroxyl radicals. The acid ionization constant of the short-lived HO transient is difficult to determine by conventional methods but an estimate can be made because HO, but not its conjugate base, O -, oxidizes ferrocyanide ions HO + Fe(CN) — OH- + Fe(CN)g . Use the following kinetic data26 for the apparent second-order rate constant as a function of pH to estimate Ka for the acid dissociation equilibrium HO + H20 = [Pg.271]

Another use for this solvent is exemplified by 1,4,5,8-tetraazanaph-thalene, the anhydrous species of which has a predicted i Ka value of — 2.7 (the observed pA in water is + 2.51). The spectrum obtained in anhydrous dichloroacetic acid is almost identical with that of the predominantly anhydrous neutral species determined in water, but quite different from the spectrum measured in dilute aqueous acid. Moreover, addition of water to the anhydrous dichloroacetic acid solution of this base caused the fine structure present in the spectrum of the neutral species to disappear and the band due to the hydrated cation (i.e. the spectrum obtained in water at pH 0.5) to appear. Addition of water to dichloroacetic acid solutions has been used to show that the cations of 3- and 8-nitro-l,6-naphthyridine20 are hydrated in aqueous acid at pH 0.5. [Pg.12]

For the tetrapeptide Gly-Gly-X—Ala, the pAa values were determined by NMR spectroscopy. They are 3.9 for Asp, 4.3 for Glu, 10.3 for Tyr and 11.1 for Lys. The p.Ka value of the C-terminal carboxyl group is approximately 3.3 and varies slightly when the amino acid in position X is exchanged. Since only the pXa values of the side chains of Glu and His are close to pH 5.0 and 5.4 used in their experiments , specific pH effects could only be expected for these two residues. [Pg.185]

As you approach the problem, you need to think about what it is you are calculating. The solution is 0.10 M acetic acid solution. That means that 0.10 mol acetic acid was dissolved in enough water to make 1.00 liters of solution. Some of that 0.10 mol has dissociated, however. Because we don t know how much, we have to determine the amount using the expression for Ka. This will allow us to determine the concentrations, and eventually the pH. We will begin by setting up a chart like the ones you used in the last chapter [Pg.325]

Step 5 Use an equilibrium table to find the H.O concentration in a weak acid or the OH concentration in a weak base. Alternatively, if the concentrations of conjugate acid and base calculated in step 4 are both large relative to the concentration of hydronium ions, use them in the expression for /<, or the Henderson—Hasselbalch equation to determine the pH. In each case, if the pH is less than 6 or greater than 8, assume that the autoprotolysis of water does not significantly affect the pH. If necessary, convert between Ka and Kh by using Kw = KA X Kb. [Pg.579]

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