Conductivity of Electrolyte Solutions

2. To titrate a hydrochloric acid solution of “known” concentration with standardized 0.5M sodium hydroxide.

From the graph, R2 can be extrapolated from the y-axis and the corresponding value on the x-axis (the point of intersection on the curve) gives the concentration of the unknown solution.

(0.074 mol HCl x 1 mol NaOH) / 1 mol HCl = 0.074 mol NaOH

The concentration or molarity of a solution can be seen through the ratio of solute amount to solution volume. This means that when you take the solute amount of a solution in moles and divide it by its volumes in liters you are able to find the solution’s overall concentration or molarity. This is proven in my data table below labeled, Molarity Simulation Data, and it helps to show that when you get the moles of a solute and the volume of the solvent you can easily and accurately figure out the molarity of the solution you are investigating. For example, when there were .75 moles of an energy drink power, and only .25 liters of water to mix it in, as shown in #3 of my data table, I was able to identify the molarity as 3 M through the process of division The way I executed this experiment lets me to see the impressionable relationship between the solute amount in moles and the solution volume in liters and how they affect molarity of a solution. The data table allows me to demonstrate this pattern. My argument is, the molarity of an aqueous solution can be found if you take the moles found in a given solute and divide them by the volume of the solvent. The data presented in this paper proves that the given formula shows the dependent, mathematical relationship that all three of these factors

If one had weighed out precisely 0.500 g of KIO3 for the primary standard solution and dissolved it in enough

The following question refers to the following: The solubility of silver phosphate (Ag3PO4) at 25oC is 1.63e–5 mol/L.

9. How many moles of NaOH would be needed to completely react with all of the excess HCl determined in problem 8?

. The second term in first equations ([Ch2E-15]) and ([Ch2E-16]) becomes much smaller than the first term, because the detuning \Delta=\omega-\omega_{\textrm{A}}\ll\omega_{\textrm{A}}

1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 Concentration (mM) y = 0.1897x R2 = 0.99

Ka is the acid dissociation constant and [HA] is the concentration of the weak acid . Strong acids usually completely dissociate and has a Ka value greater than 1. Weak acids don’t dissociate completely and have a Ka value much smaller than 1. pKa values are often used for weak acids due to being able to work with whole numbers

species in solution—or more specifically, the concentration (c) and the path length (b). the ratio of

Our purpose of doing this lab was to prove the Hess’s law correct. Hess’s law suggests that the enthalpy change of a reaction must be equal to the sum of the enthalpy changes of the related reactions which lead to the original reactions. The following are the reactions at the lab;

The maximum concentration of sediment calculated using equation 9 is compared to the concentration of sediment in the reach at the beginning of the time step, concsed, ch, i . If

This equation resulted in being left with a negative final percentage error. Which was easy to fix, as it just had to be turned into a positive for the final answer. It is evident to see how the theoretical results increase each time approximately -620 KJ/mol. This is as each carbon chain increases so does the amount of bonds associated with it.

5. In reaction three, the number of moles of NaOH can be calculated from the concentration of the solution (1.0M = 1.0mole/L) and the volume used. The calculation is below. Enter the result into Data Table 2.