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Hello everyone.
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So the question says here that we have potassium iodide that undergoes titration, potassium made having volume 25 ml, and concentration is 0 .105 molar, it undergo titration with silver nitrate having concentration of 0 .05, 1 to 4 molar.
00:21
So we have to calculate here, psg plus that we have to calculate here provided we have been given with.
00:30
Solubility product of silver iodide which is 8 .3 multiplied by 10 to minus 17.
00:37
We can say the titrant here the titrant here is your hgno3 and the analyte analyte is your potassium iodide that is ki so silver nitrate undergoes reaction with potassium iodide forming silver iodide plus potassium nitrate.
01:25
So as we can see from the reaction that one mole of silver nitrate is reacting with one mole of potassium iodide.
01:34
So one mole of silver nitrate reacting with, reacting with one mole of potassium iodide here.
01:52
One mole of potassium iodide.
01:54
So now let's find out the moles of k i present here.
01:58
Because we have been given with concentration and volume of potassium added that means we can easily calculate the number of moles of k i present in the reaction so moles of k i that will be equal to concentration multiplied by volume your concentration is 0 .105 molar multiplied by volume that is 25 and on calculating in calculating this, we will get the number of millimoles of number of millimoles of ki present in the reaction, that is 2 .515 millimoles.
02:52
Then similarly, moles of silver nitrate will be moles of silver nitrate.
03:00
It will be concentration multiplied by volume...