EDTA Standardization Data from titrations: The above EDTA...

EDTA Standardization Data from titrations: The above EDTA solution was standardized against exactly 0.2545 g calcium carbonate dissolved in exactly 250 mL and as described in the protocol. A 25 mL aliquot of this calcium carbonate solution was then titrated, and the following was found: Trial | Initial vol (mL) | final vol (mL) | total Vol | M EDTA | mmol EDTA | Find: M of EDTA and mmol of EDTA. 1 | 0.25 | 27.65 | 27.4 | ? | ? 2 | 0.9 | 28.4 | 27.3 | ? | ? 3 | 0 | 27.75 | 27.75 | ? | ? mean 27.48 standard devia 0

Transcript

00:01 Hello, in this question we are given here, that is for c .a.

00:05 Co3, calcium carbonate, mass and volume it is given here.

00:09 And we need to find out the value of the, we need to find value of molarity of editor.

00:23 So here let us first find out the number of moles of calcium carbonate.

00:27 So as we know that, that is moles of c .a.

00:32 Co3.

00:34 We know that that is it is equal to number of moles is equal to mass divided by molar mass.

00:39 So therefore on putting the value here, that is 0 .2545 is equal to molar mass of calcium carbonate.

00:48 It is 100 and we will have 0 .002545.

00:53 Number of moles now as we know that that is molarity is equal to number of moles divided by volume so here number of moles we have 0 .002545 divided by volume here it is 250 ml or 0 .250 liter it will become 0 .018m now here as we know that that is the molarity of c .o3 it is we have just calculated.

01:31 Now volume for c .c .o .3.

01:34 Volume of this calcium carbonate now it is given, that is 25mm.

01:40 So this will become 0 .025 liter.

01:43 So therefore on solving, we can find out the number of moles of c .c .o .3 by doing this titration process.

01:51 So we will have molarity into volume.

01:54 So that will become 0 .0118 multiply by 0 .025.

01:59 And that will equal to 0 .000 -02545moles.

02:09 Now on titration with adetha.

02:12 Adetita is ethylene diamine tetraacetic acid.

02:17 So here the moles of, we can say that, that is moles of adita is equal to moles of calcium carbonate, that is equal to 0 .002545 moles it is there or we can write it as 2 .545 multiply by 10 days to the power minus 4 moles...

Question

Please give Ace some feedback