Calculate the pH of each of the following strong acid solutions: (b) 0.225 g of HClO3 in 2.00 L of solution

Video transcript

Hello. Everyone in this video, we're dealing with our periodic acid, which is H. I. 04 right here. So we know that periodic acid is a strong acid. What that means is able to associate completely into its ionic components, mainly what we're focusing on. We're dealing with ph is going to be our H plus ions. So let's first start off by going ahead to write the chemical equation for this. So again, we're starting off with our periodic acid and that's gonna go ahead and associate into our H plus ions as well as R. O. H. And ions. Alright, so the information that we're given is that we have our 0.455g of our periodic acid Over 785 million years of water. I'll just note that as H20. To make it a little bit easier for us. I want to first go ahead and change the volume of milliliters units in two liters. So for every one liter of water that gives us 1000 mL of water. So you can see that the millions of HBO will cancel out nicely. Next thing I want to go ahead and do is use the molar mass of our periodic acid to go ahead and get the moles of our periodic acid. So to cancel out my grounds of periodic acid, I want to go ahead and put those units on the bottom and one mole on top. So my molar mass of periodic acid is 91.91 g of periodic acid for every one mole of periodic acid. We can see here now that the the grams of pure acid will go ahead and cancel nicely now to use the geometry and the given um chemical equation we can go ahead and get the moles of H plus ions. And we can see here that everything is a 1 to 1 ratio. So that's just one mole of p arctic acid for every one mole of H plus cat ions. Again, we can see that the moles of pr to acid will go ahead and cancel out beautifully. Now putting all of these numerical values into my calculator. I see that I get a value of 0. 025847 Molars. How I get the unit of molars here is because we have moles over L and that's the unit of concentration. So to rewrite that I have that my concentration of my H plus is equal to 0. moller. So now we can use this value to go ahead and software ph of the solution. So the equation for P. H. Is of course negative log of the concentration of R. H plus ions, which is what we just saw for. So going ahead to plug that value into my calculator. So let's see we have 0.0030-0-5847. So putting that into my calculator, I will get the value for p h two B 2.5 to 0. So this right here is going to be my final answer for this problem. Thank you all so much for watching.