(a) How many moles of analyte are present in a 10.0-μM solution that occupies 1.0 % of the lengt

(a) How many moles of analyte are present in a 10.0-μM...

Key Concepts

Electrophoretic Mobility

Electrophoretic mobility is a measure of how fast an ion moves through a medium under the influence of an electric field. It is used to link the applied voltage, the capillary length, and the time required for sample injection. Knowing the mobility allows one to calculate the necessary electric field strength to achieve the desired migration in a given timeframe.

Voltage and Electric Field Relationship

This concept pertains to the relationship between the applied voltage and the resulting electric field within the capillary. The field is determined by dividing the voltage by the capillary length. By setting the migration distance equal to the product of electrophoretic mobility, electric field, and time, one can solve for the voltage needed to drive the migration over a specified time period, while also accounting for factors such as differences in conductivity.

Mole Calculation

This concept refers to determining the number of moles present by multiplying the analyte’s molar concentration (in moles per liter) by the volume of the solution (converted to liters). It requires proper unit conversion so that the dimensions of volume and concentration are consistent, yielding the total moles present in the sample.

Capillary Geometry and Injection Volume

Understanding the geometry of the capillary is essential in establishing the volume of sample injected. This involves calculating the volume from the given dimensions and the fraction of the capillary length used for injection. Correct handling of units and spatial dimensions is key to converting these measurements into a usable volume.

Transcript

0:00 Welcome.

00:01 We are going to be discussing and doing some calculations for ion concentrations.

00:06 In each of these problems, we're going to be given how much volume.

00:12 We're going to be given volume.

00:19 And we're going to also be given either grams, which we will then convert to moles.

00:24 So i'm going to put that here.

00:26 Or moles.

00:26 Either way, we're going to have to get to moles.

00:30 We are going to be asked to solve for how many, or i shouldn't say how many.

00:36 What is the concentration? so we'll be doing this similarity.

00:44 And the way that this problem puts it is concentration of all ions present.

00:54 All ions.

00:57 So i will report that appropriately.

01:00 We're told that each one of the substances we're going to be dealing with is a strong electrolyte, which means it dissociates completely.

01:09 So these are going to be simple calculations.

01:12 Okay, for our first problem, we are given the following.

01:17 We are given that we are dealing with calcium nitrate, which has one calcium and two nitrate ions.

01:28 So that's our substance, and we have 0 .1010 moles in 100 .0, not 1 ,000, 100 .0 mill liters of solution.

01:52 Okay, so the first thing we're going to do, and i should have finished up here, we're given these, i'll do it down here.

02:01 We're going to find the molarity, malarity of the compound, the molarity of the calcium nitrate, and malarity is moles per liter.

02:16 So for our first problem, we're going to take our moles, 0 .1010 moles, and divide that by 0 .1 .000 liters.

02:27 Don't forget to turn this into liters.

02:31 And as you can see, quite simply, the concentration of this is 1 .00 molar.

02:41 And that's c .a.

02:46 N .3.

02:51 Okay.

02:52 Now, we know if we have 1 .00 molar c .a.

02:59 N .32, i could multiply this by a factor of 1 .032.

03:07 C .a.

03:08 N .o .32 contains 1 .c .a2 plus equals 1 .00 molar c .a2 plus.

03:20 I can do the same calculation, but i'm just going to this time put two nitrates for one calcium nitrate equals 2 molar.

03:45 N -o -3 plus.

03:49 So now we were asked to find the concentration of all ions present.

03:53 So the concentration of calcium was 1 .00.

03:57 The concentration of the nitrate was 2.

04:00 Let's do our second problem.

04:02 I'm going to check my camera and it's good.

04:04 Our second problem, we are given 2 .5 moles of sodium sulfate.

04:16 And this is in 1 .25 liters of solution.

04:25 So our molarity is moles per liter, is 2 .5 moles divided by 1 .25 liters, and that will be 2 .0 molarity, na2 -s .4.

04:45 Now in na2 -s -o -4, we can see that we have two sodiums, and one sulfate...

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