Learn about Junior Chemistry, Solutions 8, in this comprehensive video by bannanaiscool.
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Male Speaker: Just in case you do some labs at the great ten or 11 level that involves titration. Here are some of the apparatus that you would use. This is a retort stand here attached to it a burette clamp to hold a 50 milliliter burette with the zero line starts up here, 50 milliliters down here and you deliver a solution in this case some water into a receptacle down below and in this case the beaker or for better -- flask. Solution stoichiometry, I love like the stoichiometry that we were doing before in the other units. Okay, so here is your equation in this case, sulphuric acid and potassium hydroxide, they come to get it double replacement for potassium sulphate and h positive goes with OH negative to form HOH, as a liquid excess of water. When you balance it two here and two here no is that right now, I fly through that kind of thing and you have to be able to do but remember to write the whole equation and balance appropriately. If you don't balance appropriately you actually get the stoichiometry wrong, you have said that before. Now in the question we have got 0.15 mole per liter sulphuric acid and we are titrating it like in that demonstration, 20 mls of it with KOH, which is being poured in and it takes 30 milliliters of KOH to completely neutralize it to make an indicator change color. Why will indicators change color and what they are all, what's that all about, that's seen you love chemistry. Right now all you have to know is that there is 30 milliliters that is required to neutralize 20 here so what concentration of KOH did we need. Well, we need to find moles here do a ratio and then use the volume to find the concentration. So, we flush that out with one line of stoichiometry, okay, we have 20 milliliters now look at the way I wrote it, I wrote it as 0.0200 liters just put it in the stoichiometry as liters right away but kept my significant digits very important. So, 0.0200 liters times 0.15 moles of H2SO4 per liter what is that that cancels liters and leaves you with moles of H2SO4, don't want moles of H2SO4 want moles of KOH, so there is the ratio that taken into consideration, two moles of KOH to one mole of H2SO4, with that H2SO4 in the bottom and it cancels out there now you got moles of KOH, don't want moles and what concentration, concentration is moles divided by volume of KOH when volumes is 30 mls turn it one two three and 0.0300 liters when you do the math you are going to keep there is three significant digits 2, 3, infinite for those ones so we are going to keep two significant digits and get 0.02 moles per liter of KOH, that's how we do solutions stoichiometry.