The deviations from the mean are:. a) The standard error of the mean is the standard deviation of the mean and is given by the standard deviation of the data set divided by the square root of the number. b) The coefficient of variation is the percentage relative standard deviation or (s x/ ) × 100%. The standard error of a mean, sm, is the standard deviation of the set of data, s, divided by the square root of the number of data in the set, i.e. from the equation for the volume of a sphere that we have. deviation from the diameter.
The differences may arise due to the different algorithms used to calculate the SD and to rounding errors in the calculations. The standard error of the mean of 5 measurements is therefore smaller than the standard deviation of a single result. Since the F value calculated in the table exceeds F critical, we reject H0 and conclude that the phosphorus contents of the soil samples taken from the 3 locations are different.
Since the calculated F exceeds the critical F, we reject the null hypothesis and conclude that the mean ascorbic acid content of the 5 brands of orange juice differs at the 95% confidence level. Hydrochloric acid, HCl, is an example of a strong acid. i) Le Châtelier's principle states that the position of an equilibrium always shifts in such a direction as to relieve the stress. A common ion such as sulfate added to a solution containing sparingly soluble BaSO4 is an example. j) The common ion effect is responsible for the reduced solubility of an ionic precipitate when one of the soluble components that reacts to form the precipitate is added to the solution in equilibrium with the precipitate.
For a pure solid, the concentration of the chemical species in the solid is constant. For a given ionic strength, the activity coefficient becomes smaller as the charge of the chemical species increases. There is no change in the charge states of the ions present in the solution equilibria.
No approximations needed, because we have 4 equations and 4 unkowns
Mixed crystal formation is also a form of coprecipitation in which an impurity ion replaces an ion in the crystal lattice. Precipitation due to particle growth results in a smaller number of large particles. a) Digestion is a process in which a precipitate is heated in the presence of the solution from which it is formed (the mother liquor). Because the concentration of the impurity in the new solution is lower, the second precipitate contains less coprecipitated impurity.
A local excess of reagent does not occur and the resulting solid product is more suitable for analysis than the precipitate formed by direct addition of the precipitation reagent. e) The counterion layer describes a layer of solution containing enough excess negative ions that surrounds a charged particle. In peptization, the coagulated colloid returns to its original dispersed state due to a decrease in the concentration of the electrolyte in the solution in contact with the precipitate. The concentration of the sample is then determined from the amounts of reagent and sample, the concentration of the reagent, and the stoichiometry of the reaction.
However, a secondary standard solution of the reagent is easily prepared by standardizing a solution of sodium hydroxide against a primary standard reagent such as potassium hydrogen phthalate. mmol Fe2+ titrated back No. mL Hg mmol Hg mmol. 3 mmol Cl mmol KCl·MgCl. mL mmol KCl·MgCl mmol K L L. The eye has limited sensitivity. To see the color change requires an approximately tenfold excess of one form or the other of the indicator.
This change corresponds to a pH range of the indicator pKa ± 1 pH unit, a total range of 2 pH units. The point sharpness is affected by the completeness of the reaction between the analyte and the titrant and the concentrations of the analyte and the titrant. a) The initial pH of the NH3 solution will be lower than that of the solution containing NaOH. With the first addition of titrant, the pH of the NH3 solution will drop rapidly and then stabilize and become nearly constant during the middle part of the titration.
In contrast, additions of standard acid to the NaOH solution will cause the pH of the NaOH solution to decrease gradually and almost linearly until the equivalence point is approached. In any case, we initially assume that [H3O+] and [OH−] are much smaller than the molar concentration of the acid and the conjugate base, so that [A–] ≅ cNaA. These assumptions then lead to the following: In each of the parts of this problem we are dealing with a weak base B and its conjugate acid BHCl or (BH)2SO4.
In order to get a better picture of the change in pH with dilution, we will abandon the usual approximations and write down. Note the very small change in pH that occurs when diluting this buffer solution. g) Continuing as in part (f), a 10-fold dilution of this solution results in a pH change less than 1 to the third decimal place.
Considerations similar to those in Problem 14-41 lead to the following quadratic equation for titrating a weak base with HCl. This titration of a weak acid with a strong base follows the same basic spreadsheet as Pb 14-41, with the concentrations changed. This is a titration of a weak base with HCl and therefore follows the format of Pb 14-42.
